<?xml version='1.0' encoding='UTF-8'?><?xml-stylesheet href="http://www.blogger.com/styles/atom.css" type="text/css"?><feed xmlns='http://www.w3.org/2005/Atom' xmlns:openSearch='http://a9.com/-/spec/opensearchrss/1.0/' xmlns:georss='http://www.georss.org/georss' xmlns:gd='http://schemas.google.com/g/2005' xmlns:thr='http://purl.org/syndication/thread/1.0'><id>tag:blogger.com,1999:blog-3270928583371147820</id><updated>2012-02-16T03:51:16.520-08:00</updated><category term='IITK C SECTION'/><category term='TEXAS CSc  2001'/><category term='TEX INSTRUMENTS PAPER'/><category term='TECHNICAL TEST 2003'/><category term='TEXAS PAPER'/><category term='TEXAS IISc'/><category term='REC 2000'/><category term='IITB'/><category term='REC TRICHY 2000'/><category term='TEXAS 1999'/><title type='text'>TEXAS PAPERS</title><subtitle type='html'></subtitle><link rel='http://schemas.google.com/g/2005#feed' type='application/atom+xml' href='http://texaspapers.blogspot.com/feeds/posts/default'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default?max-results=100'/><link rel='alternate' type='text/html' href='http://texaspapers.blogspot.com/'/><link rel='hub' href='http://pubsubhubbub.appspot.com/'/><author><name>Giri Prasad Mutta</name><uri>http://www.blogger.com/profile/08563801841591925686</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='32' height='28' src='http://bp3.blogger.com/_4g1eV2MpL68/R6xyM7ZeYNI/AAAAAAAAA2o/iMoUGokYxco/S220/giri+with+pulss.JPG'/></author><generator version='7.00' uri='http://www.blogger.com'>Blogger</generator><openSearch:totalResults>11</openSearch:totalResults><openSearch:startIndex>1</openSearch:startIndex><openSearch:itemsPerPage>100</openSearch:itemsPerPage><entry><id>tag:blogger.com,1999:blog-3270928583371147820.post-8784358407161574686</id><published>2007-05-05T09:02:00.000-07:00</published><updated>2007-05-05T09:03:22.045-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='TEX INSTRUMENTS PAPER'/><title type='text'>TEXAS INSTRUMENTS PAPER</title><content type='html'>&lt;span style="font-weight:bold;"&gt;TEXAS INSTRUMENTS: TECHNICAL TEST &lt;/span&gt;&lt;br /&gt;&lt;br /&gt;**********CLICK ON THE IMAGE TO MAXIMIZE&lt;br /&gt;&lt;br /&gt;Date: 20th December 2003&lt;br /&gt;Venue: TAJ LANDS END, BANDRA-WEST, MUMBAI&lt;br /&gt;Note: Out of 100 candidates only 7 were short-listed)&lt;br /&gt;&lt;br /&gt;1. For a CMOS inverter, the transition slope of Vout vs Vin DC characteristics can be increased (steeper transition) by:&lt;br /&gt;&lt;br /&gt;a. Increasing W/L of PMOS transistor &lt;br /&gt;b. Increasing W/L of NMOS transistor &lt;br /&gt;c. Increasing W/L of both transistors by the same factor&lt;br /&gt;d. Decreasing W/L of both transistor by the same factor&lt;br /&gt;&lt;br /&gt;2. Minimum number of 2-input NAND gates that will be required to implement the function:  Y = AB + CD + EF is&lt;br /&gt;a. 4&lt;br /&gt;b. 5&lt;br /&gt;c. 6&lt;br /&gt;d. 7&lt;br /&gt;&lt;br /&gt;3. Consider a two-level memory hierarchy system M1 &amp; M2. M1 is accessed first and on miss M2 is accessed. The access of M1 is 2 nanoseconds and the miss penalty (the time to get the data from M2 in case of a miss) is 100 nanoseconds. The probability that a valid data is found in M1 is 0.97. The average memory access time is:&lt;br /&gt;a. 4.94 nanoseconds&lt;br /&gt;b. 3.06 nanoseconds&lt;br /&gt;c. 5.00 nanoseconds&lt;br /&gt;d. 5.06 nanoseconds&lt;br /&gt;&lt;br /&gt;4. Interrupt latency is the time elapsed between:&lt;br /&gt;a. Occurrence of an interrupt and its detection by the CPU&lt;br /&gt;b. Assertion of an interrupt and the start of the associated ISR&lt;br /&gt;c. Assertion of an interrupt and the completion of the associated ISR&lt;br /&gt;d. Start and completion of associated ISR&lt;br /&gt;&lt;br /&gt;5. Which of the following is true for the function (A.B + A’.C + B.C)&lt;br /&gt;a. This function can glitch and can be further reduced&lt;br /&gt;b. This function can neither glitch nor can be further reduced&lt;br /&gt;c. This function can glitch and cannot be further reduced&lt;br /&gt;d. This function cannot glitch but can be further reduced&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_hRx4l03TQT8/RjjxOWx4wBI/AAAAAAAAAsc/pHYgAVwLT2g/s1600-h/6.JPG"&gt;&lt;img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://4.bp.blogspot.com/_hRx4l03TQT8/RjjxOWx4wBI/AAAAAAAAAsc/pHYgAVwLT2g/s400/6.JPG" border="0" alt=""id="BLOGGER_PHOTO_ID_5060059410286034962" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;br /&gt;8. A CPU supports 250 instructions. Each instruction op-code has these fields:&lt;br /&gt;• The instruction type (one among 250)&lt;br /&gt;• A conditional register specification&lt;br /&gt;• 3 register operands&lt;br /&gt;• Addressing mode specification for both source operands&lt;br /&gt;&lt;br /&gt;The CPU has 16 registers and supports 5 addressing modes. What is the instruction op-code length in bits?&lt;br /&gt;a. 32&lt;br /&gt;b. 24&lt;br /&gt;c. 30&lt;br /&gt;d. 36&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_hRx4l03TQT8/Rjjx_Gx4wCI/AAAAAAAAAsk/X7L04uOO1tE/s1600-h/7.JPG"&gt;&lt;img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://3.bp.blogspot.com/_hRx4l03TQT8/Rjjx_Gx4wCI/AAAAAAAAAsk/X7L04uOO1tE/s400/7.JPG" border="0" alt=""id="BLOGGER_PHOTO_ID_5060060247804657698" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_hRx4l03TQT8/RjjyRmx4wDI/AAAAAAAAAss/sTiaeVhc1mY/s1600-h/8.JPG"&gt;&lt;img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://1.bp.blogspot.com/_hRx4l03TQT8/RjjyRmx4wDI/AAAAAAAAAss/sTiaeVhc1mY/s400/8.JPG" border="0" alt=""id="BLOGGER_PHOTO_ID_5060060565632237618" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;br /&gt;12. Which of the following statements is/are true?&lt;br /&gt;I. Combinational circuits may have feedback, sequential circuits do not.&lt;br /&gt;II. Combinational circuits have a ‘memory-less’ property, sequential circuits do not.&lt;br /&gt;III. Both combinational and sequential circuits must be controlled by an external clock.&lt;br /&gt;&lt;br /&gt;a. I only&lt;br /&gt;b. II and III only&lt;br /&gt;c. I and II only&lt;br /&gt;d. II only&lt;br /&gt;&lt;br /&gt;13. Consider an alternate binary number representation scheme, wherein the number of ones M, in a word of N bits, is always the same. This scheme is called the M-out-of-N coding scheme. If M=N/2, and N=8, what is the efficiency of this coding scheme as against the regular binary number representation scheme? (As a hint, consider that the number of unique words represent able in the latter representation with N bits is 2^N. Hence the efficiency is 100%)&lt;br /&gt;a. Close to 30%&lt;br /&gt;b. Close to 50%&lt;br /&gt;c. Close to 70%&lt;br /&gt;d. Close to 100%&lt;br /&gt;&lt;br /&gt;14. A CPU supports 4 interrupts- I1, I2, I3 and I4. It supports priority of interrupts. Nested interrupts are allowed if later interrupt is higher priority than previous one. During a certain period of time, we observe the following sequence of entry into and exit from the interrupt service routine:&lt;br /&gt;I1-start---I2-start---I2-end---I4-start---I3-start---I3-end---I4-end---I1-end&lt;br /&gt;From this sequence, what can we infer about the interrupt routines?&lt;br /&gt;a. I3 &gt; I4 &gt; I2 &gt; I1&lt;br /&gt;b. I4 &gt; I3 &gt; I2 &gt; I1&lt;br /&gt;c. I2 &gt; I1; I3 &gt; I4 &gt; I1&lt;br /&gt;d. I2 &gt; I1, I3 &gt; I4 &gt; I2 &gt; I1&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;15. I decide to build myself a small electric kettle to boil my cup of tea. I need 200 ml of water for my cup of tea. Assuming that typical tap water temperature is 25 C and I want the water boiling in exactly one minute, then what is the wattage required for the heating element?&lt;br /&gt;[Assume: Boiling point of water is 100 C, 1 Calorie (heat required to change 1 gm of water by 1 C)= 4 joules, 1 ml of water weighs 1 gm.]&lt;br /&gt;a. Data given is insufficient&lt;br /&gt;b. 800 W&lt;br /&gt;c. 300 W&lt;br /&gt;d. 1000 W&lt;br /&gt;e. 250 W&lt;br /&gt;&lt;br /&gt;16. The athletics team from REC Trichy is traveling by train. The train slows down, (but does not halt) at a small wayside station that has  a 100 mts long platform. The sprinter (who can run 100 mts in 10 sec) decides to jump down and get a newspaper and some idlis. He jumps out just as his compartment enters the platform and spends 5 secs buying his newspaper that is at the point where he jumped out. He then sprints along the platform to buy idlis that is another 50 mts. He spends another 5 secs  buying the idlis. He is now  just 50 mts from the other end of the platform where the train is moving out. He begins running in the direction of the train and the only other open door in his train is located 50 mts behind the door from where he jumped. At what(uniform) speed should the train be traveled if he just misses jumping into the open door at the very edge of the platform?&lt;br /&gt;Make the following assumptions&lt;br /&gt;• He always runs at his peak speed uniformly&lt;br /&gt;• The train travels at uniform speed&lt;br /&gt;• He does not wait (other than for the idlis &amp; newspaper) or run baclwards&lt;br /&gt;&lt;br /&gt;a. Data given is insufficient&lt;br /&gt;b. 4 m/s&lt;br /&gt;c. 5 m/s&lt;br /&gt;d. 7.5 m/s&lt;br /&gt;e. 10 m/s&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;17. State which of the following gate combinations does not form a universal logic set:&lt;br /&gt;a. 2-input AND + 2-input OR&lt;br /&gt;b. 2-to-1 multiplexer&lt;br /&gt;c. 2-input XOR + inverter&lt;br /&gt;d. 3-input NAND&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_hRx4l03TQT8/Rjjy1Gx4wEI/AAAAAAAAAs0/UQYp9YwL39w/s1600-h/9.JPG"&gt;&lt;img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://3.bp.blogspot.com/_hRx4l03TQT8/Rjjy1Gx4wEI/AAAAAAAAAs0/UQYp9YwL39w/s400/9.JPG" border="0" alt=""id="BLOGGER_PHOTO_ID_5060061175517593666" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;br /&gt;19. The FSM (finite state machine) below starts in state Sa, which is the reset state, and detects a particular sequence of inputs leading it to state Sc. FSMs have a few characteristics. An autonomous FSM has no inputs. For a Moore FSM, the output depends on the present state alone. For a Mealy FSM, the output depends on the present state as well as the inputs. Which of the statements best describes the FSM below?&lt;br /&gt;&lt;br /&gt;a. It has two states and is autonomous&lt;br /&gt;b. The information available is insufficient&lt;br /&gt;c. It is a Mealy machine with three states&lt;br /&gt;d. It is a Moor machine with three states&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_hRx4l03TQT8/RjjzLWx4wFI/AAAAAAAAAs8/dqvY0M3Tku4/s1600-h/1.JPG"&gt;&lt;img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://4.bp.blogspot.com/_hRx4l03TQT8/RjjzLWx4wFI/AAAAAAAAAs8/dqvY0M3Tku4/s400/1.JPG" border="0" alt=""id="BLOGGER_PHOTO_ID_5060061557769683026" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_hRx4l03TQT8/RjjzZGx4wGI/AAAAAAAAAtE/qXTnGK-lVuo/s1600-h/2.JPG"&gt;&lt;img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://3.bp.blogspot.com/_hRx4l03TQT8/RjjzZGx4wGI/AAAAAAAAAtE/qXTnGK-lVuo/s400/2.JPG" border="0" alt=""id="BLOGGER_PHOTO_ID_5060061793992884322" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;br /&gt;22. The value (0xdeadbeef) needs to stored at address 0x400. Which of the below ways will the memory look like in a big endian machine:&lt;br /&gt;&lt;br /&gt; 0x403   0x402   0x401   0x400&lt;br /&gt;a.  be   ef   de   ad&lt;br /&gt;b. ef   be   ad   de&lt;br /&gt;c. fe   eb   da   ed&lt;br /&gt;d. ed   da   eb   fe&lt;br /&gt;&lt;br /&gt;23. In a given CPU-memory sub-system, all accesses to the memory take two cycles. Accesses to memories in two consecutive cycles can therefore result in incorrect data transfer. Which of the following access mechanisms guarantees correct data transfer?&lt;br /&gt;a. A read operation followed by a write operation in the next cycle.&lt;br /&gt;b. A write operation followed by a read operation in the next cycle.&lt;br /&gt;c. A NOP between every successive reads &amp; writes&lt;br /&gt;d. None of the above&lt;br /&gt;&lt;br /&gt;24. An architecture saves 4 control registers automatically on function entry (and restores them on function return). Save of each registers costs 1 cycle (so does restore). How many cycles are spent in these tasks (save and restore) while running the following un-optimized code with n=5:&lt;br /&gt;&lt;br /&gt;Void fib(int n)&lt;br /&gt;{&lt;br /&gt; if((n==0) || (n==1)) return 1;&lt;br /&gt; return(fib(n-1) + fib(n-2));&lt;br /&gt;}&lt;br /&gt;a. 120&lt;br /&gt;b. 80&lt;br /&gt;c. 125&lt;br /&gt;d. 128&lt;br /&gt;&lt;br /&gt;25. The maximum number of unique Boolean functions F(A,B), realizable for a two input (A,B) and single output (Z) circuit is:&lt;br /&gt;&lt;br /&gt;a. 2&lt;br /&gt;b. 6&lt;br /&gt;c. 80&lt;br /&gt;d. None of the above&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_hRx4l03TQT8/Rjj0CWx4wHI/AAAAAAAAAtM/WGxfyytgWZ0/s1600-h/4.JPG"&gt;&lt;img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://4.bp.blogspot.com/_hRx4l03TQT8/Rjj0CWx4wHI/AAAAAAAAAtM/WGxfyytgWZ0/s400/4.JPG" border="0" alt=""id="BLOGGER_PHOTO_ID_5060062502662488178" /&gt;&lt;/a&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3270928583371147820-8784358407161574686?l=texaspapers.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://texaspapers.blogspot.com/feeds/8784358407161574686/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=3270928583371147820&amp;postID=8784358407161574686' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/8784358407161574686'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/8784358407161574686'/><link rel='alternate' type='text/html' href='http://texaspapers.blogspot.com/2007/05/texas-instruments-paper.html' title='TEXAS INSTRUMENTS PAPER'/><author><name>Giri Prasad Mutta</name><uri>http://www.blogger.com/profile/08563801841591925686</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='32' height='28' src='http://bp3.blogger.com/_4g1eV2MpL68/R6xyM7ZeYNI/AAAAAAAAA2o/iMoUGokYxco/S220/giri+with+pulss.JPG'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://4.bp.blogspot.com/_hRx4l03TQT8/RjjxOWx4wBI/AAAAAAAAAsc/pHYgAVwLT2g/s72-c/6.JPG' height='72' width='72'/><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3270928583371147820.post-1694324901148640492</id><published>2007-05-05T08:50:00.000-07:00</published><updated>2007-05-05T08:51:15.682-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='TECHNICAL TEST 2003'/><title type='text'>TEXAS TECHNICAL TEST 2003</title><content type='html'>&lt;span style="font-weight:bold;"&gt;TEXAS INSTRUMENTS: TECHNICAL TEST &lt;/span&gt;&lt;br /&gt;&lt;br /&gt;**********CLICK ON THE IMAGE TO MAXIMIZE&lt;br /&gt;&lt;br /&gt;Date: 20th December 2003&lt;br /&gt;Venue: TAJ LANDS END, BANDRA-WEST, MUMBAI&lt;br /&gt;Note: Out of 100 candidates only 7 were short-listed)&lt;br /&gt;&lt;br /&gt;1. For a CMOS inverter, the transition slope of Vout vs Vin DC characteristics can be increased (steeper transition) by:&lt;br /&gt;&lt;br /&gt;a. Increasing W/L of PMOS transistor &lt;br /&gt;b. Increasing W/L of NMOS transistor &lt;br /&gt;c. Increasing W/L of both transistors by the same factor&lt;br /&gt;d. Decreasing W/L of both transistor by the same factor&lt;br /&gt;&lt;br /&gt;2. Minimum number of 2-input NAND gates that will be required to implement the function:  Y = AB + CD + EF is&lt;br /&gt;a. 4&lt;br /&gt;b. 5&lt;br /&gt;c. 6&lt;br /&gt;d. 7&lt;br /&gt;&lt;br /&gt;3. Consider a two-level memory hierarchy system M1 &amp; M2. M1 is accessed first and on miss M2 is accessed. The access of M1 is 2 nanoseconds and the miss penalty (the time to get the data from M2 in case of a miss) is 100 nanoseconds. The probability that a valid data is found in M1 is 0.97. The average memory access time is:&lt;br /&gt;a. 4.94 nanoseconds&lt;br /&gt;b. 3.06 nanoseconds&lt;br /&gt;c. 5.00 nanoseconds&lt;br /&gt;d. 5.06 nanoseconds&lt;br /&gt;&lt;br /&gt;4. Interrupt latency is the time elapsed between:&lt;br /&gt;a. Occurrence of an interrupt and its detection by the CPU&lt;br /&gt;b. Assertion of an interrupt and the start of the associated ISR&lt;br /&gt;c. Assertion of an interrupt and the completion of the associated ISR&lt;br /&gt;d. Start and completion of associated ISR&lt;br /&gt;&lt;br /&gt;5. Which of the following is true for the function (A.B + A’.C + B.C)&lt;br /&gt;a. This function can glitch and can be further reduced&lt;br /&gt;b. This function can neither glitch nor can be further reduced&lt;br /&gt;c. This function can glitch and cannot be further reduced&lt;br /&gt;d. This function cannot glitch but can be further reduced&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_hRx4l03TQT8/RjjxOWx4wBI/AAAAAAAAAsc/pHYgAVwLT2g/s1600-h/6.JPG"&gt;&lt;img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://4.bp.blogspot.com/_hRx4l03TQT8/RjjxOWx4wBI/AAAAAAAAAsc/pHYgAVwLT2g/s400/6.JPG" border="0" alt=""id="BLOGGER_PHOTO_ID_5060059410286034962" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;br /&gt;8. A CPU supports 250 instructions. Each instruction op-code has these fields:&lt;br /&gt;• The instruction type (one among 250)&lt;br /&gt;• A conditional register specification&lt;br /&gt;• 3 register operands&lt;br /&gt;• Addressing mode specification for both source operands&lt;br /&gt;&lt;br /&gt;The CPU has 16 registers and supports 5 addressing modes. What is the instruction op-code length in bits?&lt;br /&gt;a. 32&lt;br /&gt;b. 24&lt;br /&gt;c. 30&lt;br /&gt;d. 36&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_hRx4l03TQT8/Rjjx_Gx4wCI/AAAAAAAAAsk/X7L04uOO1tE/s1600-h/7.JPG"&gt;&lt;img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://3.bp.blogspot.com/_hRx4l03TQT8/Rjjx_Gx4wCI/AAAAAAAAAsk/X7L04uOO1tE/s400/7.JPG" border="0" alt=""id="BLOGGER_PHOTO_ID_5060060247804657698" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_hRx4l03TQT8/RjjyRmx4wDI/AAAAAAAAAss/sTiaeVhc1mY/s1600-h/8.JPG"&gt;&lt;img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://1.bp.blogspot.com/_hRx4l03TQT8/RjjyRmx4wDI/AAAAAAAAAss/sTiaeVhc1mY/s400/8.JPG" border="0" alt=""id="BLOGGER_PHOTO_ID_5060060565632237618" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;br /&gt;12. Which of the following statements is/are true?&lt;br /&gt;I. Combinational circuits may have feedback, sequential circuits do not.&lt;br /&gt;II. Combinational circuits have a ‘memory-less’ property, sequential circuits do not.&lt;br /&gt;III. Both combinational and sequential circuits must be controlled by an external clock.&lt;br /&gt;&lt;br /&gt;a. I only&lt;br /&gt;b. II and III only&lt;br /&gt;c. I and II only&lt;br /&gt;d. II only&lt;br /&gt;&lt;br /&gt;13. Consider an alternate binary number representation scheme, wherein the number of ones M, in a word of N bits, is always the same. This scheme is called the M-out-of-N coding scheme. If M=N/2, and N=8, what is the efficiency of this coding scheme as against the regular binary number representation scheme? (As a hint, consider that the number of unique words represent able in the latter representation with N bits is 2^N. Hence the efficiency is 100%)&lt;br /&gt;a. Close to 30%&lt;br /&gt;b. Close to 50%&lt;br /&gt;c. Close to 70%&lt;br /&gt;d. Close to 100%&lt;br /&gt;&lt;br /&gt;14. A CPU supports 4 interrupts- I1, I2, I3 and I4. It supports priority of interrupts. Nested interrupts are allowed if later interrupt is higher priority than previous one. During a certain period of time, we observe the following sequence of entry into and exit from the interrupt service routine:&lt;br /&gt;I1-start---I2-start---I2-end---I4-start---I3-start---I3-end---I4-end---I1-end&lt;br /&gt;From this sequence, what can we infer about the interrupt routines?&lt;br /&gt;a. I3 &gt; I4 &gt; I2 &gt; I1&lt;br /&gt;b. I4 &gt; I3 &gt; I2 &gt; I1&lt;br /&gt;c. I2 &gt; I1; I3 &gt; I4 &gt; I1&lt;br /&gt;d. I2 &gt; I1, I3 &gt; I4 &gt; I2 &gt; I1&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;15. I decide to build myself a small electric kettle to boil my cup of tea. I need 200 ml of water for my cup of tea. Assuming that typical tap water temperature is 25 C and I want the water boiling in exactly one minute, then what is the wattage required for the heating element?&lt;br /&gt;[Assume: Boiling point of water is 100 C, 1 Calorie (heat required to change 1 gm of water by 1 C)= 4 joules, 1 ml of water weighs 1 gm.]&lt;br /&gt;a. Data given is insufficient&lt;br /&gt;b. 800 W&lt;br /&gt;c. 300 W&lt;br /&gt;d. 1000 W&lt;br /&gt;e. 250 W&lt;br /&gt;&lt;br /&gt;16. The athletics team from REC Trichy is traveling by train. The train slows down, (but does not halt) at a small wayside station that has  a 100 mts long platform. The sprinter (who can run 100 mts in 10 sec) decides to jump down and get a newspaper and some idlis. He jumps out just as his compartment enters the platform and spends 5 secs buying his newspaper that is at the point where he jumped out. He then sprints along the platform to buy idlis that is another 50 mts. He spends another 5 secs  buying the idlis. He is now  just 50 mts from the other end of the platform where the train is moving out. He begins running in the direction of the train and the only other open door in his train is located 50 mts behind the door from where he jumped. At what(uniform) speed should the train be traveled if he just misses jumping into the open door at the very edge of the platform?&lt;br /&gt;Make the following assumptions&lt;br /&gt;• He always runs at his peak speed uniformly&lt;br /&gt;• The train travels at uniform speed&lt;br /&gt;• He does not wait (other than for the idlis &amp; newspaper) or run baclwards&lt;br /&gt;&lt;br /&gt;a. Data given is insufficient&lt;br /&gt;b. 4 m/s&lt;br /&gt;c. 5 m/s&lt;br /&gt;d. 7.5 m/s&lt;br /&gt;e. 10 m/s&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;17. State which of the following gate combinations does not form a universal logic set:&lt;br /&gt;a. 2-input AND + 2-input OR&lt;br /&gt;b. 2-to-1 multiplexer&lt;br /&gt;c. 2-input XOR + inverter&lt;br /&gt;d. 3-input NAND&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_hRx4l03TQT8/Rjjy1Gx4wEI/AAAAAAAAAs0/UQYp9YwL39w/s1600-h/9.JPG"&gt;&lt;img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://3.bp.blogspot.com/_hRx4l03TQT8/Rjjy1Gx4wEI/AAAAAAAAAs0/UQYp9YwL39w/s400/9.JPG" border="0" alt=""id="BLOGGER_PHOTO_ID_5060061175517593666" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;br /&gt;19. The FSM (finite state machine) below starts in state Sa, which is the reset state, and detects a particular sequence of inputs leading it to state Sc. FSMs have a few characteristics. An autonomous FSM has no inputs. For a Moore FSM, the output depends on the present state alone. For a Mealy FSM, the output depends on the present state as well as the inputs. Which of the statements best describes the FSM below?&lt;br /&gt;&lt;br /&gt;a. It has two states and is autonomous&lt;br /&gt;b. The information available is insufficient&lt;br /&gt;c. It is a Mealy machine with three states&lt;br /&gt;d. It is a Moor machine with three states&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_hRx4l03TQT8/RjjzLWx4wFI/AAAAAAAAAs8/dqvY0M3Tku4/s1600-h/1.JPG"&gt;&lt;img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://4.bp.blogspot.com/_hRx4l03TQT8/RjjzLWx4wFI/AAAAAAAAAs8/dqvY0M3Tku4/s400/1.JPG" border="0" alt=""id="BLOGGER_PHOTO_ID_5060061557769683026" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_hRx4l03TQT8/RjjzZGx4wGI/AAAAAAAAAtE/qXTnGK-lVuo/s1600-h/2.JPG"&gt;&lt;img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://3.bp.blogspot.com/_hRx4l03TQT8/RjjzZGx4wGI/AAAAAAAAAtE/qXTnGK-lVuo/s400/2.JPG" border="0" alt=""id="BLOGGER_PHOTO_ID_5060061793992884322" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;br /&gt;22. The value (0xdeadbeef) needs to stored at address 0x400. Which of the below ways will the memory look like in a big endian machine:&lt;br /&gt;&lt;br /&gt; 0x403   0x402   0x401   0x400&lt;br /&gt;a.  be   ef   de   ad&lt;br /&gt;b. ef   be   ad   de&lt;br /&gt;c. fe   eb   da   ed&lt;br /&gt;d. ed   da   eb   fe&lt;br /&gt;&lt;br /&gt;23. In a given CPU-memory sub-system, all accesses to the memory take two cycles. Accesses to memories in two consecutive cycles can therefore result in incorrect data transfer. Which of the following access mechanisms guarantees correct data transfer?&lt;br /&gt;a. A read operation followed by a write operation in the next cycle.&lt;br /&gt;b. A write operation followed by a read operation in the next cycle.&lt;br /&gt;c. A NOP between every successive reads &amp; writes&lt;br /&gt;d. None of the above&lt;br /&gt;&lt;br /&gt;24. An architecture saves 4 control registers automatically on function entry (and restores them on function return). Save of each registers costs 1 cycle (so does restore). How many cycles are spent in these tasks (save and restore) while running the following un-optimized code with n=5:&lt;br /&gt;&lt;br /&gt;Void fib(int n)&lt;br /&gt;{&lt;br /&gt; if((n==0) || (n==1)) return 1;&lt;br /&gt; return(fib(n-1) + fib(n-2));&lt;br /&gt;}&lt;br /&gt;a. 120&lt;br /&gt;b. 80&lt;br /&gt;c. 125&lt;br /&gt;d. 128&lt;br /&gt;&lt;br /&gt;25. The maximum number of unique Boolean functions F(A,B), realizable for a two input (A,B) and single output (Z) circuit is:&lt;br /&gt;&lt;br /&gt;a. 2&lt;br /&gt;b. 6&lt;br /&gt;c. 80&lt;br /&gt;d. None of the above&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_hRx4l03TQT8/Rjj0CWx4wHI/AAAAAAAAAtM/WGxfyytgWZ0/s1600-h/4.JPG"&gt;&lt;img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://4.bp.blogspot.com/_hRx4l03TQT8/Rjj0CWx4wHI/AAAAAAAAAtM/WGxfyytgWZ0/s400/4.JPG" border="0" alt=""id="BLOGGER_PHOTO_ID_5060062502662488178" /&gt;&lt;/a&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3270928583371147820-1694324901148640492?l=texaspapers.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://texaspapers.blogspot.com/feeds/1694324901148640492/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=3270928583371147820&amp;postID=1694324901148640492' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/1694324901148640492'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/1694324901148640492'/><link rel='alternate' type='text/html' href='http://texaspapers.blogspot.com/2007/05/texas-technical-test-2003.html' title='TEXAS TECHNICAL TEST 2003'/><author><name>Giri Prasad Mutta</name><uri>http://www.blogger.com/profile/08563801841591925686</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='32' height='28' src='http://bp3.blogger.com/_4g1eV2MpL68/R6xyM7ZeYNI/AAAAAAAAA2o/iMoUGokYxco/S220/giri+with+pulss.JPG'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://4.bp.blogspot.com/_hRx4l03TQT8/RjjxOWx4wBI/AAAAAAAAAsc/pHYgAVwLT2g/s72-c/6.JPG' height='72' width='72'/><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3270928583371147820.post-5127163368473240507</id><published>2007-05-05T08:49:00.002-07:00</published><updated>2007-05-05T08:50:44.186-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='TEXAS 1999'/><title type='text'>TEXAS 1999</title><content type='html'>Here is Texas paper for you.&lt;br /&gt;    in this paper there was 20 questions as follows in 60 minutes .&lt;br /&gt;    second part consists of 36 que. in 30 minutes all questions are&lt;br /&gt;    diagramatical.(figurs)..&lt;br /&gt;&lt;br /&gt;    1. if a 5-stage pipe-line is flushed and then we have to execute 5 and&lt;br /&gt;12&lt;br /&gt;    instructions respectively then no. of  cycles will be&lt;br /&gt;    a. 5 and 12&lt;br /&gt;    b. 6 and 13&lt;br /&gt;    c. 9 and 16&lt;br /&gt;    d.none&lt;br /&gt;&lt;br /&gt;    2. k-map&lt;br /&gt;&lt;br /&gt;    ab&lt;br /&gt;    ----------&lt;br /&gt;    c       1   x  0  0&lt;br /&gt;           1   x  0  x&lt;br /&gt;&lt;br /&gt;    solve it&lt;br /&gt;&lt;br /&gt;    a. A.B&lt;br /&gt;    B. ~A&lt;br /&gt;    C. ~B&lt;br /&gt;    D. A+B&lt;br /&gt;&lt;br /&gt;    3.CHAR  A[10][15] AND INT B[10][15] IS DEFINED&lt;br /&gt;    WHAT'S THE ADDRESS OF A[3][4] AND B[3][4]&lt;br /&gt;    IF ADDRESS OD A IS OX1000 AND B IS 0X2000&lt;br /&gt;&lt;br /&gt;    A. 0X1030 AND 0X20C3&lt;br /&gt;    B. OX1031 AND OX20C4&lt;br /&gt;    AND SOME OTHERS..&lt;br /&gt;&lt;br /&gt;    4. int f(int *a)&lt;br /&gt;    {&lt;br /&gt;    int b=5;&lt;br /&gt;    a=&amp;b;&lt;br /&gt;    }&lt;br /&gt;&lt;br /&gt;    main()&lt;br /&gt;    {&lt;br /&gt;    int i;&lt;br /&gt;    printf("\n %d",i);&lt;br /&gt;    f(&amp;i);&lt;br /&gt;    printf("\n %d",i);&lt;br /&gt;    }&lt;br /&gt;&lt;br /&gt;    what's the output .&lt;br /&gt;&lt;br /&gt;    1.10,5&lt;br /&gt;    2,10,10&lt;br /&gt;    c.5,5&lt;br /&gt;    d. none&lt;br /&gt;&lt;br /&gt;    5. main()&lt;br /&gt;    {&lt;br /&gt;    int i;&lt;br /&gt;    fork();&lt;br /&gt;    fork();&lt;br /&gt;    fork();&lt;br /&gt;    printf("----");&lt;br /&gt;    }&lt;br /&gt;&lt;br /&gt;    how many times the printf will be executed .&lt;br /&gt;    a.3&lt;br /&gt;    b. 6&lt;br /&gt;    c.5&lt;br /&gt;    d. 8&lt;br /&gt;&lt;br /&gt;    6.&lt;br /&gt;    void f(int i)&lt;br /&gt;    {&lt;br /&gt;    int j;&lt;br /&gt;    for (j=0;j&lt;16;j++)&lt;br /&gt;    {&lt;br /&gt;    if (i &amp; (0x8000&gt;&gt;j))&lt;br /&gt;    printf("1");&lt;br /&gt;    else&lt;br /&gt;    printf("0");&lt;br /&gt;    }&lt;br /&gt;    }&lt;br /&gt;    what's the purpose of the program&lt;br /&gt;&lt;br /&gt;    a. its output is hex representation of i&lt;br /&gt;    b. bcd&lt;br /&gt;    c. binary&lt;br /&gt;    d. decimal&lt;br /&gt;&lt;br /&gt;    7.#define f(a,b) a+b&lt;br /&gt;    #define g(a,b) a*b&lt;br /&gt;&lt;br /&gt;    main()&lt;br /&gt;    {&lt;br /&gt;&lt;br /&gt;    int m;&lt;br /&gt;    m=2*f(3,g(4,5));&lt;br /&gt;    printf("\n m is %d",m);&lt;br /&gt;    }&lt;br /&gt;&lt;br /&gt;    what's the value of m&lt;br /&gt;    a.70&lt;br /&gt;    b.50&lt;br /&gt;    c.26&lt;br /&gt;    d. 69&lt;br /&gt;&lt;br /&gt;    8.&lt;br /&gt;    main()&lt;br /&gt;    {&lt;br /&gt;    char a[10];&lt;br /&gt;    strcpy(a,"\0");&lt;br /&gt;    if (a==NULL)&lt;br /&gt;    printf("\a is null");&lt;br /&gt;    else&lt;br /&gt;    printf("\n a is not null");}&lt;br /&gt;&lt;br /&gt;    what happens with it .&lt;br /&gt;    a. compile time error.&lt;br /&gt;    b. run-time error.&lt;br /&gt;    c. a is  null&lt;br /&gt;    d. a is not  null.&lt;br /&gt;&lt;br /&gt;    9. char a[5]="hello"&lt;br /&gt;&lt;br /&gt;    a. in array we can't do the operation .&lt;br /&gt;    b. size of a is too large&lt;br /&gt;    c. size of a is too small&lt;br /&gt;    d. nothing wrong with it .&lt;br /&gt;&lt;br /&gt;    10. local variables can be store by compiler&lt;br /&gt;    a. in register or heap&lt;br /&gt;    b. in register or stack&lt;br /&gt;    c .in stack or heap .&lt;br /&gt;    d. global memory.&lt;br /&gt;&lt;br /&gt;    11. average and worst time complexity in a sorted binary tree is&lt;br /&gt;&lt;br /&gt;    12. a tree is given and ask to find its meaning (parse-tree)&lt;br /&gt;    (expression tree)&lt;br /&gt;    ans. ((a+b)-(c*d))  ( not confirmed)&lt;br /&gt;    13. convert 40.xxxx into binary .&lt;br /&gt;&lt;br /&gt;    14. global variable conflicts due to multiple file occurance&lt;br /&gt;    is resolved during&lt;br /&gt;    a. compile-time&lt;br /&gt;    b. run-time&lt;br /&gt;    c. link-time&lt;br /&gt;    d. load-time&lt;br /&gt;&lt;br /&gt;    15.&lt;br /&gt;    two program is given of factorial.&lt;br /&gt;    one  with  recursion and one without recursion .&lt;br /&gt;    question was which program won't run for very big no. input     because&lt;br /&gt;of stack overfow .&lt;br /&gt;    a. i only  (ans.)&lt;br /&gt;    b. ii only&lt;br /&gt;    c. i&amp; ii both .&lt;br /&gt;    c. none&lt;br /&gt;&lt;br /&gt;    16.&lt;br /&gt;    struct a&lt;br /&gt;    {&lt;br /&gt;    int a;&lt;br /&gt;    char b;&lt;br /&gt;    int c;&lt;br /&gt;    }&lt;br /&gt;&lt;br /&gt;    union b&lt;br /&gt;    {&lt;br /&gt;    char a;&lt;br /&gt;    int b;&lt;br /&gt;    int c;&lt;br /&gt;    };&lt;br /&gt;    which is correct .&lt;br /&gt;    a. size of a is always diff. form size of b.(ans.)&lt;br /&gt;    b. size of a is always same  form size of b.&lt;br /&gt;    c. we can't say anything because of not-homogeneous (not in ordered)&lt;br /&gt;    d. size of a can be same if ...&lt;br /&gt;&lt;br /&gt;                                   bye..&lt;br /&gt;                                            p.sreenivasa rao&lt;br /&gt;&lt;br /&gt;______________________________________________________&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3270928583371147820-5127163368473240507?l=texaspapers.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://texaspapers.blogspot.com/feeds/5127163368473240507/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=3270928583371147820&amp;postID=5127163368473240507' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/5127163368473240507'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/5127163368473240507'/><link rel='alternate' type='text/html' href='http://texaspapers.blogspot.com/2007/05/texas-1999_05.html' title='TEXAS 1999'/><author><name>Giri Prasad Mutta</name><uri>http://www.blogger.com/profile/08563801841591925686</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='32' height='28' src='http://bp3.blogger.com/_4g1eV2MpL68/R6xyM7ZeYNI/AAAAAAAAA2o/iMoUGokYxco/S220/giri+with+pulss.JPG'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3270928583371147820.post-3809868802017970765</id><published>2007-05-05T08:49:00.001-07:00</published><updated>2007-05-05T08:49:17.401-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='IITB'/><title type='text'>TEXAS IITB</title><content type='html'>&gt;&lt;span style="font-weight:bold;"&gt; &gt; &gt; TEXAS INSTRUMENTS PAPER&lt;/span&gt;&lt;br /&gt;&gt; &gt; &gt; ------------------------&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 1.THE CURRENT IN 4-OHM RESISTOR IS (CIRCUIT IS&lt;br /&gt;&gt; &gt; &gt; GIVEN)&lt;br /&gt;&gt; &gt; &gt; ANS:1A&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 2.WHAT IS THE BELOW CIRCUIT&lt;br /&gt;&gt; &gt; &gt; A)MEMORY ELEMENT&lt;br /&gt;&gt; &gt; &gt; B)LATCH&lt;br /&gt;&gt; &gt; &gt; C)AND GATE&lt;br /&gt;&gt; &gt; &gt; D)OSCILLATOR&lt;br /&gt;&gt; &gt; &gt; ANS:NOT SURE&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 3.KARNAUGH MAP&lt;br /&gt;&gt; &gt; &gt; ANS:B'(B-BAR)&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 4.HOW MANY 2:1 MUX ARE REQUIRED TO IMPLEMENT 8:1 MUX&lt;br /&gt;&gt; &gt; &gt; ANS:7&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 5.TWO  N_MOS FETS ARE THERE .WHAT IS THE OUTPUT=20&lt;br /&gt;&gt; &gt; &gt; ANS:3VOLTS&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 6.#DEFINE SUM (A,B)  A+B&lt;br /&gt;&gt; &gt; &gt; FIND THE VALUE OF X&lt;br /&gt;&gt; &gt; &gt; X=3DSUM(2,3)*SUM(3,2)&lt;br /&gt;&gt; &gt; &gt; ANS:13(NONE OF THE ABOVE)&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 7.FIND VALUE OF SIGMA B[i] =B7(-2)POWER I&lt;br /&gt;&gt; &gt; &gt; WHERE B[i] IS THE VALUE AT THE ITH  PLACE&lt;br /&gt;&gt; &gt; &gt; WHAT IS RESULT IF INPUT IS 0 1 1 0 0  1 1 0&lt;br /&gt;&gt; &gt; &gt; ANS:+34&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 8.(A EX-OR B)EX-OR B=3D&lt;br /&gt;&gt; &gt; &gt; ANS:A&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 9.AB+CD+EF HOW MANY TWO I/P NAND GATES ARE REQUIRED&lt;br /&gt;&gt; &gt; &gt; TO IMPLEMENT THE ABOV=&lt;br /&gt;&gt; &gt; &gt; E FUNCTIION&lt;br /&gt;&gt; &gt; &gt; ANS:6&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 10.HOW MANY BOOLEAN FUNCTIONS WE CAN REALIZE WITH&lt;br /&gt;&gt; &gt; &gt; ONE INPUT&lt;br /&gt;&gt; &gt; &gt; ANS:4&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 11. WHAT IS THE TOTAL POWER DISSIPATION IF EACH BULB&lt;br /&gt;&gt; &gt; &gt; RATED AT 20 W  ,120V=&lt;br /&gt;&gt; &gt; &gt; OLTS(IN THE  GIVEN FIGURE)&lt;br /&gt;&gt; &gt; &gt; ANS:12W&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 12. TO INCREASE THE SLOPE OF C-MOS .WHAT IS TO BE&lt;br /&gt;&gt; &gt; &gt; DONE.&lt;br /&gt;&gt; &gt; &gt; ANS:INCREASE W/L RATIO OF P-MOS&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 13.IF THERE ARE W PEOPLE STANDING IN THE QUEUE AT&lt;br /&gt;&gt; &gt; &gt; FIRST IF THE RATE OF RE=&lt;br /&gt;&gt; &gt; &gt; MOVAL IS PROPORTIONAL TO REMAINING PERSONS IN THE&lt;br /&gt;&gt; &gt; &gt; QUEUE.&lt;br /&gt;&gt; &gt; &gt; ANS:DECREASES EXPONENTIALLY&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 14.SIMLIFY THE FOLLOWING BOOLEAN EXPRESSION=20&lt;br /&gt;&gt; &gt; &gt; A'+A'C+AB'&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 15.AB'+BC IN THE FOLLOWING ANSWERS WHICH ARE&lt;br /&gt;&gt; &gt; &gt; EQUILENT TO GIVEN FORM FUNCT=&lt;br /&gt;&gt; &gt; &gt; IONALLY&lt;br /&gt;&gt; &gt; &gt; ANS:ALL OF THE ABOVE&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 16.CIRCUIT CONSISTS OF 6 FETS IS GIVEN. O/P IS&lt;br /&gt;&gt; &gt; &gt; ANS:(A+BC)'&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 17. IN THE FOLLOWING IF 8 I/P 'S IS GIVEN TO AND&lt;br /&gt;&gt; &gt; &gt; GATE .WHICH IS THE FASTE=&lt;br /&gt;&gt; &gt; &gt; ST IMPLEMENTATION=20&lt;br /&gt;&gt; &gt; &gt; ANS: ((( AB)(CD))((EF)(GH)))&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 18. IF PROBABILITY OF EACH INPUT HIGH IS 0.5&lt;br /&gt;&gt; &gt; &gt; AND LOW IS 0.5.WHAT IS THE PROBABILITY OF O/P AS&lt;br /&gt;&gt; &gt; &gt; HIGH.&lt;br /&gt;&gt; &gt; &gt; CIRCUIT IS GIVEN)OUTPUT Y=3D(AB)+(CD)'&lt;br /&gt;&gt; &gt; &gt; ANS:13/16.&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 19.OUT PUT OF THE GIVEN CIRCUIT(WHEAT STONE BRIDGE&lt;br /&gt;&gt; &gt; &gt; WITH DIODES)&lt;br /&gt;&gt; &gt; &gt; ANS:0 VOLTS&lt;br /&gt;&gt; &gt; &gt;&lt;br /&gt;&gt; &gt; &gt; 20.IF C IS THE CAPACITENCE AND V IS THE APPLIED&lt;br /&gt;&gt; &gt; &gt; VOLTAGE AND F IS THE SWIT=&lt;br /&gt;&gt; &gt; &gt; HING FREQUENCY THEN POWER DISSIPATION IS&lt;br /&gt;&gt; &gt; &gt; ANS: F*C*V*V&lt;br /&gt;&gt; &gt; &gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3270928583371147820-3809868802017970765?l=texaspapers.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://texaspapers.blogspot.com/feeds/3809868802017970765/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=3270928583371147820&amp;postID=3809868802017970765' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/3809868802017970765'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/3809868802017970765'/><link rel='alternate' type='text/html' href='http://texaspapers.blogspot.com/2007/05/texas-iitb.html' title='TEXAS IITB'/><author><name>Giri Prasad Mutta</name><uri>http://www.blogger.com/profile/08563801841591925686</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='32' height='28' src='http://bp3.blogger.com/_4g1eV2MpL68/R6xyM7ZeYNI/AAAAAAAAA2o/iMoUGokYxco/S220/giri+with+pulss.JPG'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3270928583371147820.post-3737687350144988570</id><published>2007-05-05T08:48:00.003-07:00</published><updated>2007-05-05T08:48:58.642-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='REC 2000'/><title type='text'>TEXAS REC 2000</title><content type='html'>&lt;span style="font-weight:bold;"&gt;&lt;br /&gt;   M.N.R.E.C  TI PAPER   (2000)&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;   about all interviews in brief they are asking about n/w's so prepare well.&lt;br /&gt;&lt;br /&gt;      20 tech q's    (1 hr)&lt;br /&gt;    tech:&lt;br /&gt;     multiple choice , q's are not in order&lt;br /&gt;&lt;br /&gt;     1)Reentrant os required &lt;br /&gt;         a)multitasking&lt;br /&gt;         b)multiuser &lt;br /&gt;         c)both&lt;br /&gt;         d)none&lt;br /&gt;     ans: c   (check)&lt;br /&gt;  &lt;br /&gt;     2)identify a DFA for all strings which ended with string '00'&lt;br /&gt;             4 dfa's have given , u can simply identify&lt;br /&gt;     3)dfa has given we have to find the string accepted by that.&lt;br /&gt;    &lt;br /&gt;     4)in a memory hierarchy, M1,M2&lt;br /&gt;        M1 access time is 2 ns ,M2 accesstime is 100 ns . the hit ratio for &lt;br /&gt;        M1 is .97, then what's the avg access time&lt;br /&gt;      ans: 500 ns&lt;br /&gt;&lt;br /&gt;     5) Interrupts main use is for elimination of &lt;br /&gt;     a) spooling&lt;br /&gt;     b)polling&lt;br /&gt;     c)job scheduling&lt;br /&gt;     d)....&lt;br /&gt;    &lt;br /&gt;     6)A coprocessor for floating point operations has given. after that the execution of the&lt;br /&gt;       floating point ops has incresed by 20 times. if 20%  floating pt ops are there then&lt;br /&gt;        what's the sped up?&lt;br /&gt;&lt;br /&gt;     7)  two heaps of sizes m,n are given,&lt;br /&gt;    then the time needed to merge them&lt;br /&gt;    a)m&lt;br /&gt;     b)n&lt;br /&gt;    c)m+n&lt;br /&gt;    d) none&lt;br /&gt;  8)context switching is useful in&lt;br /&gt;    a) spooling&lt;br /&gt;    b) polling&lt;br /&gt;    c)interrupt handling&lt;br /&gt;    d) interrupt servicing&lt;br /&gt;&lt;br /&gt; 9)  if 2^n leaves are given in a tree then no.of internal nodes &lt;br /&gt;   a)2^n&lt;br /&gt;   b) 2^n -1&lt;br /&gt;   c);......&lt;br /&gt;&lt;br /&gt;10) semaphores used for &lt;br /&gt;  a)to prevent deadlocks&lt;br /&gt;  b) to sycronize critical activities&lt;br /&gt;  c)....&lt;br /&gt; &lt;br /&gt;  remaining q's i have forgot, but some q's are about c (easy)&lt;br /&gt;&lt;br /&gt;   u can do 18 q's in this section ( i.e this sec is that much easy)&lt;br /&gt;&lt;br /&gt;anal:&lt;br /&gt;    it's some what difficult.&lt;br /&gt;  in this sec they has given big passeges, analytical reasoning,some apti bits&lt;br /&gt;  i have forgot many of the q's&lt;br /&gt;  so  i am giving here few q's&lt;br /&gt;  but it's very easy to shortlist in this TI, if u done this sec well.&lt;br /&gt;&lt;br /&gt;1) derivative of e^sinx&lt;br /&gt;2)x=2yt y=2t^2&lt;br /&gt;   dy/dx?    (figures may not correct)&lt;br /&gt;    &lt;br /&gt;&lt;br /&gt;  x mod y=x-y[x/y]&lt;br /&gt;        =x-y{x/y}    some conditions are given&lt;br /&gt;   [x/y] means floor operation.&lt;br /&gt;   {x/y} ceal op.&lt;br /&gt;&lt;br /&gt;based on this 4 or 5 q's are given u can do the all the q's except&lt;br /&gt;  one q.&lt;br /&gt;  that's why  iam giving only one q from this &lt;br /&gt; &lt;br /&gt;3)xmod 3=2, xmod5=10 then xmod15=?&lt;br /&gt;&lt;br /&gt;  i can't give the remaining q's those are very lengthy i have forgot those.&lt;br /&gt;&lt;br /&gt;about TI interview:   &lt;br /&gt;&lt;br /&gt;  Ti is asking v.....very basic fundaasssss&lt;br /&gt;   for me they have asked only in c, but some other persi\ons they asked in ds,os,ip etc.&lt;br /&gt;&lt;br /&gt;   in c they have asked&lt;br /&gt;   if u have given a 32 bit no. u have to set m th bit without using any arithmetic ops&lt;br /&gt;   and u must have to write in a single expr. (hint: use &amp;,&lt;&lt; ops)&lt;br /&gt;   they asked the same for reset.&lt;br /&gt;  &lt;br /&gt;  they have asked to print 2^0 to 2^16 with out using any arith. ops. (same as above)&lt;br /&gt;if u have given two seperate c files u have toexecute these two in one by linking how &lt;br /&gt; u can achieve this (hint: make utility)&lt;br /&gt;&lt;br /&gt;about ds:&lt;br /&gt;    they have asked how u can print levelwise in a tree (   hint: bfs)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3270928583371147820-3737687350144988570?l=texaspapers.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://texaspapers.blogspot.com/feeds/3737687350144988570/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=3270928583371147820&amp;postID=3737687350144988570' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/3737687350144988570'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/3737687350144988570'/><link rel='alternate' type='text/html' href='http://texaspapers.blogspot.com/2007/05/texas-rec-2000.html' title='TEXAS REC 2000'/><author><name>Giri Prasad Mutta</name><uri>http://www.blogger.com/profile/08563801841591925686</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='32' height='28' src='http://bp3.blogger.com/_4g1eV2MpL68/R6xyM7ZeYNI/AAAAAAAAA2o/iMoUGokYxco/S220/giri+with+pulss.JPG'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3270928583371147820.post-6125537695570046066</id><published>2007-05-05T08:48:00.001-07:00</published><updated>2007-05-05T08:48:33.627-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='IITK C SECTION'/><title type='text'>TEXAS IITK C SECTION</title><content type='html'>Here is the C section.(not in order)&lt;br /&gt;&lt;br /&gt;1. x=x^y;&lt;br /&gt;   y=x^y;&lt;br /&gt;   x=x^y;&lt;br /&gt;x=?&lt;br /&gt;&lt;br /&gt;2. In a link list we want to go from &lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;last to first element and vice&lt;br /&gt;versa.&lt;br /&gt;Which is efficient&lt;br /&gt;a. Cirular List&lt;br /&gt;b. Doubly Link List&lt;br /&gt;c. Double Ended Link List&lt;br /&gt;d. Simple List&lt;br /&gt;&lt;br /&gt;3.Hash function h(i)=i mod 5, there are 5 buckets 0,1,2,3,4. Rehashing is&lt;br /&gt;(h(i)+1)mod 5,(h(i)+2)mod 5......&lt;br /&gt;5 numbers are to be enterd, Find the number in a given bucket..&lt;br /&gt;&lt;br /&gt;4.To reconstuct a Tree which sequence is enough&lt;br /&gt;a. Inoreder sequence&lt;br /&gt;b. Preorder and Post order&lt;br /&gt;c. In, Pre, Post order are required&lt;br /&gt;d. Any one can be used&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;6.convert a inorder string to post order&lt;br /&gt;&lt;br /&gt;7.int (*a[5])() what is this declaration?&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;There were in all 10 questions in this. Also a section on Analog Circuits,&lt;br /&gt;Digital Circuits and Computer Architecture. Do any 2 of 4.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3270928583371147820-6125537695570046066?l=texaspapers.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://texaspapers.blogspot.com/feeds/6125537695570046066/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=3270928583371147820&amp;postID=6125537695570046066' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/6125537695570046066'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/6125537695570046066'/><link rel='alternate' type='text/html' href='http://texaspapers.blogspot.com/2007/05/texas-iitk-c-section.html' title='TEXAS IITK C SECTION'/><author><name>Giri Prasad Mutta</name><uri>http://www.blogger.com/profile/08563801841591925686</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='32' height='28' src='http://bp3.blogger.com/_4g1eV2MpL68/R6xyM7ZeYNI/AAAAAAAAA2o/iMoUGokYxco/S220/giri+with+pulss.JPG'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3270928583371147820.post-7014984045435890563</id><published>2007-05-05T08:47:00.002-07:00</published><updated>2007-05-05T08:48:05.570-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='REC TRICHY 2000'/><title type='text'>TEXAS REC TRICHY 2000</title><content type='html'>Texas Instruments. rec trichy this year (2000).&lt;br /&gt;don't bank on it. might be.&lt;br /&gt;&lt;br /&gt;-----------------------------------------------------------------------&lt;br /&gt;75 qs in 60 min.....&lt;br /&gt;--   2 reading comprehesion.. 5 qs each...&lt;br /&gt;--   quanta....5 qs.&lt;br /&gt;--   data sufficiency..5 qs.&lt;br /&gt;Q..u have weights only in power of 3... based on&lt;br /&gt;it answer 5&lt;br /&gt;qs. like for given weight what are the combination of&lt;br /&gt;available&lt;br /&gt;weight..&lt;br /&gt;--   data interpretation 2 qs..5 qs each&lt;br /&gt;Q..one qs witin a circle population % is given&lt;br /&gt;based on it 5&lt;br /&gt;qs.(PI chart)&lt;br /&gt;--   analitical&lt;br /&gt;Q.. one tour is shedhuled for 9 countries..each&lt;br /&gt;country have 2&lt;br /&gt;days shedhule....tour held on for 12 days..&lt;br /&gt;some cond'n are given..like if u visit ROME u&lt;br /&gt;have to visit&lt;br /&gt;australia .....&lt;br /&gt;based on it  5 qs...like which 3 countries&lt;br /&gt;missed...&lt;br /&gt;Q..4 metals are given..each year % price of these&lt;br /&gt;metals are&lt;br /&gt;increasing..some more cond'n.....based on it 4 qs..&lt;br /&gt;answer of one  qs is 13.6%&lt;br /&gt;Q.. {x}=greater real of x&lt;br /&gt;[x]=least real of x&lt;br /&gt;x mod y = some expression is given&lt;br /&gt;based on it 5 qs...like calculate 5 mod 3....&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3270928583371147820-7014984045435890563?l=texaspapers.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://texaspapers.blogspot.com/feeds/7014984045435890563/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=3270928583371147820&amp;postID=7014984045435890563' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/7014984045435890563'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/7014984045435890563'/><link rel='alternate' type='text/html' href='http://texaspapers.blogspot.com/2007/05/texas-rec-trichy-2000.html' title='TEXAS REC TRICHY 2000'/><author><name>Giri Prasad Mutta</name><uri>http://www.blogger.com/profile/08563801841591925686</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='32' height='28' src='http://bp3.blogger.com/_4g1eV2MpL68/R6xyM7ZeYNI/AAAAAAAAA2o/iMoUGokYxco/S220/giri+with+pulss.JPG'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3270928583371147820.post-5663998631841258410</id><published>2007-05-05T08:47:00.001-07:00</published><updated>2007-05-05T08:47:37.302-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='TEXAS PAPER'/><title type='text'>TEXAS PAPER</title><content type='html'>2)  simplification of boolean function   &lt;br /&gt;2) one question on sampling frequency&lt;br /&gt;3) FSM(finite state machine): Adder: can be implimented using single FSM, Multiplier can take more than one FSM&lt;br /&gt;4) ((a*b)+c)/d  how many stack operations : 9&lt;br /&gt;5) 128k bytes of cache , 32 bytes each, total memory 128Mbytes cache will devide the memory in how many :  1024.&lt;br /&gt;6) Gate output is of the above ckt : tristate inverter with B as enable for high impedence state&lt;br /&gt;7) totla number of two input boolean logic function can be implemented : 6 (OR, AND, EXOR, EXNOR, NAND, NOR)&lt;br /&gt;&lt;br /&gt;8) y= ab+ a’c+bc     Ans: may have glitch and can be reduced&lt;br /&gt;9) which of the following is not an universal logic set , (a)2 to 1 Mux, (b)3-input NAND (c)2-EXOR+1-NOT (d)2-input AND + 2input OR.&lt;br /&gt;10) REC trychy , train problem ans : 7.5 m/s&lt;br /&gt;11) Voltage at node B for a ckt with two voltage sources of 10 v and 20 v. Ands: -5v&lt;br /&gt;   If the question is of voltage drop across R , it is ‘0’ . There is no current flowing through that element&lt;br /&gt;12) two sequential flow digrams are given of 3 and  4 state rsply. So ans. is 3 and 4 .&lt;br /&gt;13) Max frequency of operation is&lt;br /&gt;T= T(steup time) + T(delay time)+ T(Skew time)&lt;br /&gt;14) one question on big endean memory format (whether LSB is stored at starting address .)&lt;br /&gt;15) CMOS questions : Vout Vs. Vin characteristics depend on W/L ratio.(answer which of the MOSFETS are having more steeper characteristics if the W/L ratios of PMOs or Nmos or both are goibg to be increased ).    &lt;br /&gt;16)  Realeted to aptitude do all aptitude questions , if time permites then go to english setion. &lt;br /&gt;    Maintain time to answer aptitude section , because u should answer 60% of the aptitude .&lt;br /&gt;All the best&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3270928583371147820-5663998631841258410?l=texaspapers.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://texaspapers.blogspot.com/feeds/5663998631841258410/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=3270928583371147820&amp;postID=5663998631841258410' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/5663998631841258410'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/5663998631841258410'/><link rel='alternate' type='text/html' href='http://texaspapers.blogspot.com/2007/05/texas-paper.html' title='TEXAS PAPER'/><author><name>Giri Prasad Mutta</name><uri>http://www.blogger.com/profile/08563801841591925686</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='32' height='28' src='http://bp3.blogger.com/_4g1eV2MpL68/R6xyM7ZeYNI/AAAAAAAAA2o/iMoUGokYxco/S220/giri+with+pulss.JPG'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3270928583371147820.post-8322450619965100919</id><published>2007-05-05T08:46:00.001-07:00</published><updated>2007-05-05T08:46:47.682-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='TEXAS 1999'/><title type='text'>TEXAS 1999</title><content type='html'>----------------------------------------------------------------------------&lt;br /&gt;&lt;br /&gt;                                                     &gt;^M&lt;br /&gt;&gt;          THIS IS &lt;span style="font-weight:bold;"&gt;TI 1999 jadavpur&lt;/span&gt; for ECE students.for cs another paper is ^M&lt;br /&gt;&gt;given^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;1.two transistors are connected Vbe is 0.7volts .this is simple ckt.one ^M&lt;br /&gt;&gt;transistor is diode equivalent. &amp; asked the o/p across the 2 nd transistor.^M&lt;br /&gt;&gt;2.simple k map ans is Bbar.^M&lt;br /&gt;&gt;3.^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;                               Emitter^M&lt;br /&gt;&gt;---R-------transistorbase| --^M&lt;br /&gt;&gt;                          | ---^M&lt;br /&gt;&gt;                                   collector^M&lt;br /&gt;&gt;             in above capacitor is connected parallel with resistance ^M&lt;br /&gt;&gt;r.capacitor is not shown^M&lt;br /&gt;&gt;             in fig.capacitor is used for in this ckt:^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;             ans:a.speedupb.active bypass  c.decoupling^M&lt;br /&gt;&gt;   4.^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;   -----R------I----------o/p^M&lt;br /&gt;&gt;           |___R____ |^M&lt;br /&gt;&gt;                             in above r is resistence.I is cmos inverter.^M&lt;br /&gt;&gt;                             then ckt is used for:^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;                             a.schmitt trigger b.latch  c.inverter  ^M&lt;br /&gt;&gt;d.amplifier^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;      5.simple amplifier ckt  openloop gain of amplifier is 4.V in ^M&lt;br /&gt;&gt;=1v.asked for V x?^M&lt;br /&gt;&gt;      amplifdier + is connected to base. - is connected to i/p in between ^M&lt;br /&gt;&gt;5k is connected.^M&lt;br /&gt;&gt;      from o/p feedback connected to - of amplifier with 15k.this is ckt.^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;      6.resistence inductot cap are serially connected to ac voltage 5 ^M&lt;br /&gt;&gt;volts.voltage across^M&lt;br /&gt;&gt;      inductor is given.R I C values are given &amp; asked for^M&lt;br /&gt;&gt;      voltages across resistence &amp; capacitor.^M&lt;br /&gt;&gt;      7.^M&lt;br /&gt;&gt;              ___  R_____^M&lt;br /&gt;&gt;             |            |^M&lt;br /&gt;&gt;      ---R------OPAMP ----------^M&lt;br /&gt;&lt;br /&gt;                 &gt;             |---^M&lt;br /&gt;&gt;             R1        R1 is for wjhat i mean what is the purpose of R1.^M&lt;br /&gt;&gt;             |^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;             ground^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;     8.asked for Vo at the o/p.it is like simple cmos realization that is n ^M&lt;br /&gt;&gt;block is above^M&lt;br /&gt;&gt;     &amp; p block is below.Vdd is 3 volts at supply.V threshold 5 volts.^M&lt;br /&gt;&gt;     9.2 d ffs are connected in asyncro manner .clock 10 MEGAHZ.gate delay ^M&lt;br /&gt;&gt;is 1 nanosec.^M&lt;br /&gt;&gt;     A B are the two given D FFs.asked for AB output is:^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;     a.updown^M&lt;br /&gt;&gt;     b.up c. updown glitching like that (take care abt glitching word)^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;     10.^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;     ----------------| subtractor|---------o/p^M&lt;br /&gt;&gt;         |___HPF____|^M&lt;br /&gt;&lt;br /&gt;                         &gt;^M&lt;br /&gt;&gt;                     the ckt is LPF ,HPF or APF ?^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;   11.in a queue at the no of elements removed is proportional to no of ^M&lt;br /&gt;&gt;elements in^M&lt;br /&gt;&gt;   the queue.then no of elements in the queue:^M&lt;br /&gt;&gt;   a.increases decreases exp or linearly(so these are the 4 options given ^M&lt;br /&gt;&gt;choose 1 option)^M&lt;br /&gt;&gt;   12.with 2 i/p AND gates u have to form a 8 i/p AND gate.which is the ^M&lt;br /&gt;&gt;fastest in the^M&lt;br /&gt;&gt;   following implementations.^M&lt;br /&gt;&gt;   ans we think ((AB)(CD))((EF)(GH))^M&lt;br /&gt;&gt;   13.with howmany 2:1 MUX u can for   8:1 MUX.answer is 7.^M&lt;br /&gt;&gt;   14. there are n states then ffs used are log n.^M&lt;br /&gt;&gt;   15.cube each side has r units resistence then the resistence across ^M&lt;br /&gt;&gt;diagonal of cube.^M&lt;br /&gt;&gt;   16.op amp connections asked for o/p^M&lt;br /&gt;&gt;   the answer is (1+1/n)(v2-v1).check it out.practise this type of model.^M&lt;br /&gt;&gt;   17.^M&lt;br /&gt;&gt;       _____________ supply^M&lt;br /&gt;&gt;   ---|__           ___|^M&lt;br /&gt;&gt;  Ii     &gt;________ |___    Tranistot^M&lt;br /&gt;&gt;                       &gt; _______Vo^M&lt;br /&gt;&gt;                       &gt; _______Vo^M&lt;br /&gt;&gt;                        |^M&lt;br /&gt;&gt;                        |^M&lt;br /&gt;&gt;                        R       |^M&lt;br /&gt;&gt;                        |       |  Io^M&lt;br /&gt;&gt;                        ground.^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;    asked for Io/Ii=? transistor gain is beta.^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;    a.(1+beta)square b.1+beta  c. beta^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;    18.y=kxsquare. this is transfer function of a block with i/p x &amp; o/p ^M&lt;br /&gt;&gt;y.if i/p is^M&lt;br /&gt;&gt;    sum of a &amp; b then o/p is :--^M&lt;br /&gt;&gt;^M&lt;br /&gt;&gt;    a. AM b.FM  c. PM^M&lt;br /&gt;&gt;    19.^M&lt;br /&gt;&gt;                 ------MULTIPLIER--- |^M&lt;br /&gt;&lt;br /&gt;                       &gt;                |                    |^M&lt;br /&gt;&gt;        _____R__|__OPAMP______________________Vo^M&lt;br /&gt;&gt;                 ---^M&lt;br /&gt;&gt;                 |^M&lt;br /&gt;&gt;                ground.^M&lt;br /&gt;&gt;                v in = -Ez then o/p Vo =?^M&lt;br /&gt;&gt;                answer is squareroot of -Ez.multiplier i/ps are a &amp; b then ^M&lt;br /&gt;&gt;its o/p^M&lt;br /&gt;&gt;                is a.b;^M&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3270928583371147820-8322450619965100919?l=texaspapers.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://texaspapers.blogspot.com/feeds/8322450619965100919/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=3270928583371147820&amp;postID=8322450619965100919' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/8322450619965100919'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/8322450619965100919'/><link rel='alternate' type='text/html' href='http://texaspapers.blogspot.com/2007/05/texas-1999.html' title='TEXAS 1999'/><author><name>Giri Prasad Mutta</name><uri>http://www.blogger.com/profile/08563801841591925686</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='32' height='28' src='http://bp3.blogger.com/_4g1eV2MpL68/R6xyM7ZeYNI/AAAAAAAAA2o/iMoUGokYxco/S220/giri+with+pulss.JPG'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3270928583371147820.post-2656194023527949135</id><published>2007-05-05T08:45:00.002-07:00</published><updated>2007-05-05T08:46:19.230-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='TEXAS IISc'/><title type='text'>TEXAS IISc</title><content type='html'>Technical was "very easy" but Aptitide was lengthy &amp; hence&lt;br /&gt;difficult.&lt;br /&gt;Interview was cool...........&lt;br /&gt;      Just clear the test.............and get it.&lt;br /&gt;&lt;br /&gt;      I appeared for CS sections, there were easy programs on C and in&lt;br /&gt;other section few questions on DSA, OS and Architecture.&lt;br /&gt;&lt;br /&gt;      1. binary search complexity&lt;br /&gt;      2. for a paging system 2 programs were given u have to choose the&lt;br /&gt;faster one&lt;br /&gt;      3. 1 q. on bigendian&lt;br /&gt;      4. If high priority process  comes when low prority process is&lt;br /&gt;running, what will OS do?&lt;br /&gt;      5. i simple qn on Hash table with rehashing.&lt;br /&gt;            5 nos were given with rehashing , find the location at&lt;br /&gt;2nd position.&lt;br /&gt;      6. difference between combinational and sequential logic&lt;br /&gt;      7. find formula for effective access time calculation... hit&lt;br /&gt;ratio,cache access time and memory access time was given.&lt;br /&gt;      8. int *ptr[4]() something like this.........&lt;br /&gt;            what does it represent?&lt;br /&gt;      9. In an architecture at each fun. call it stores 4 registers and&lt;br /&gt;on return loads the 4. each load and store take 1 cycle.&lt;br /&gt;A recursive fn to find fibonacci seriese is given with start val  as 5.&lt;br /&gt;Calculate the num of extra cycles reqd in load/store.&lt;br /&gt;&lt;br /&gt;n = 5;&lt;br /&gt;fib(int n)&lt;br /&gt;{      if(n ==0 || n==1)&lt;br /&gt;            return(1);&lt;br /&gt;      return( fib(n-1) + fib(n-2) );&lt;br /&gt;}&lt;br /&gt;ans 120    check.           &lt;br /&gt;&lt;br /&gt;&lt;br /&gt;      Can't recall all the questions ..............sorry...............&lt;br /&gt;      Apti is imp........... prepare for that............&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3270928583371147820-2656194023527949135?l=texaspapers.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://texaspapers.blogspot.com/feeds/2656194023527949135/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=3270928583371147820&amp;postID=2656194023527949135' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/2656194023527949135'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/2656194023527949135'/><link rel='alternate' type='text/html' href='http://texaspapers.blogspot.com/2007/05/texas-iisc.html' title='TEXAS IISc'/><author><name>Giri Prasad Mutta</name><uri>http://www.blogger.com/profile/08563801841591925686</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='32' height='28' src='http://bp3.blogger.com/_4g1eV2MpL68/R6xyM7ZeYNI/AAAAAAAAA2o/iMoUGokYxco/S220/giri+with+pulss.JPG'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3270928583371147820.post-4068486017081692291</id><published>2007-05-05T08:45:00.001-07:00</published><updated>2007-05-05T08:45:46.842-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='TEXAS CSc  2001'/><title type='text'>TEXAS CSc 2001</title><content type='html'>hi all&lt;br /&gt;&lt;span style="font-weight:bold;"&gt;i am sending you TI CSE paper 2001&lt;/span&gt;&lt;br /&gt;ENJOY.&lt;br /&gt;BEST OF LUCK TO ALL OF YOU.&lt;br /&gt;&lt;br /&gt;    about all interviews in brief they are asking&lt;br /&gt;about n/w's so prepare well.&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight:bold;"&gt;       20 tech q's    (1 hr)&lt;/span&gt;&lt;br /&gt;     tech:&lt;br /&gt;      multiple choice , q's are not in order&lt;br /&gt;&lt;br /&gt;      1)Reentrant os required&lt;br /&gt;          a)multitasking&lt;br /&gt;          b)multiuser&lt;br /&gt;          c)both&lt;br /&gt;          d)none&lt;br /&gt;      ans: c   (check)&lt;br /&gt;  &lt;br /&gt;      2)identify a DFA for all strings which ended&lt;br /&gt;with string '00'&lt;br /&gt;              4 dfa's have given , u can simply&lt;br /&gt;identify&lt;br /&gt;      3)dfa has given we have to find the string&lt;br /&gt;accepted by that.&lt;br /&gt;    &lt;br /&gt;      4)in a memory hierarchy, M1,M2&lt;br /&gt;         M1 access time is 2 ns ,M2 accesstime is 100&lt;br /&gt;ns . the hit ratio for&lt;br /&gt;         M1 is .97, then what's the avg access time&lt;br /&gt;       ans: 500 ns&lt;br /&gt;&lt;br /&gt;      5) Interrupts main use is for elimination of&lt;br /&gt;      a) spooling&lt;br /&gt;      b)polling&lt;br /&gt;      c)job scheduling&lt;br /&gt;      d)....&lt;br /&gt;    &lt;br /&gt;      6)A coprocessor for floating point operations&lt;br /&gt;has given. after that the execution of the&lt;br /&gt;        floating point ops has incresed by 20 times.&lt;br /&gt;if 20%  floating pt ops are there then&lt;br /&gt;         what's the sped up?&lt;br /&gt;&lt;br /&gt;      7)  two heaps of sizes m,n are given,&lt;br /&gt;     then the time needed to merge them&lt;br /&gt;     a)m&lt;br /&gt;      b)n&lt;br /&gt;     c)m+n&lt;br /&gt;     d) none&lt;br /&gt;   8)context switching is useful in&lt;br /&gt;     a) spooling&lt;br /&gt;     b) polling&lt;br /&gt;     c)interrupt handling&lt;br /&gt;     d) interrupt servicing&lt;br /&gt;&lt;br /&gt;  9)  if 2^n leaves are given in a tree then no.of&lt;br /&gt;internal nodes&lt;br /&gt;    a)2^n&lt;br /&gt;    b) 2^n -1&lt;br /&gt;    c);......&lt;br /&gt;&lt;br /&gt;10) semaphores used for&lt;br /&gt;   a)to prevent deadlocks&lt;br /&gt;   b) to sycronize critical activities&lt;br /&gt;   c)....&lt;br /&gt; &lt;br /&gt;   remaining q's i have forgot, but some q's are&lt;br /&gt;about c (easy)&lt;br /&gt;&lt;br /&gt;    u can do 18 q's in this section ( i.e this sec is&lt;br /&gt;that much easy)&lt;br /&gt;&lt;br /&gt;anal:&lt;br /&gt;     it's some what difficult.&lt;br /&gt;   in this sec they has given big passeges,&lt;br /&gt;analytical reasoning,some apti bits&lt;br /&gt;   i have forgot many of the q's&lt;br /&gt;   so  i am giving here few q's&lt;br /&gt;   but it's very easy to shortlist in this TI, if u&lt;br /&gt;done this sec well.&lt;br /&gt;&lt;br /&gt;1) derivative of e^sinx&lt;br /&gt;2)x=2yt y=2t^2&lt;br /&gt;    dy/dx?    (figures may not correct)&lt;br /&gt;    &lt;br /&gt;&lt;br /&gt;   x mod y=x-y[x/y]&lt;br /&gt;         =x-y{x/y}    some conditions are given&lt;br /&gt;    [x/y] means floor operation.&lt;br /&gt;    {x/y} ceal op.&lt;br /&gt;&lt;br /&gt;based on this 4 or 5 q's are given u can do the all&lt;br /&gt;the q's except&lt;br /&gt;   one q.&lt;br /&gt;   that's why  iam giving only one q from this&lt;br /&gt; &lt;br /&gt;3)xmod 3=2, xmod5=10 then xmod15=?&lt;br /&gt;&lt;br /&gt;   i can't give the remaining q's those are very&lt;br /&gt;lengthy i have forgot those.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;about TI interview:&lt;br /&gt;   &lt;br /&gt;   Ti is asking v.....very basic fundaasssss&lt;br /&gt;    for me they have asked only in c, but some other&lt;br /&gt;persi\ons they asked in ds,os,ip etc.&lt;br /&gt;&lt;br /&gt;    in c they have asked&lt;br /&gt;    if u have given a 32 bit no. u have to set m th&lt;br /&gt;bit without using any arithmetic ops&lt;br /&gt;    and u must have to write in a single expr. (hint:&lt;br /&gt;use &amp;,&lt;&lt; ops)&lt;br /&gt;    they asked the same for reset.&lt;br /&gt;  &lt;br /&gt;   they have asked to print 2^0 to 2^16 with out&lt;br /&gt;using any arith. ops. (same as above)&lt;br /&gt;if u have given two seperate c files u have&lt;br /&gt;toexecute these two in one by linking how&lt;br /&gt;  u can achieve this (hint: make utility)&lt;br /&gt;&lt;br /&gt;about ds:&lt;br /&gt;     they have asked how u can print levelwise in a&lt;br /&gt;tree (   hint: bfs)&lt;br /&gt;*******************************************&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3270928583371147820-4068486017081692291?l=texaspapers.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://texaspapers.blogspot.com/feeds/4068486017081692291/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=3270928583371147820&amp;postID=4068486017081692291' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/4068486017081692291'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3270928583371147820/posts/default/4068486017081692291'/><link rel='alternate' type='text/html' href='http://texaspapers.blogspot.com/2007/05/texas-csc-2001.html' title='TEXAS CSc 2001'/><author><name>Giri Prasad Mutta</name><uri>http://www.blogger.com/profile/08563801841591925686</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='32' height='28' src='http://bp3.blogger.com/_4g1eV2MpL68/R6xyM7ZeYNI/AAAAAAAAA2o/iMoUGokYxco/S220/giri+with+pulss.JPG'/></author><thr:total>0</thr:total></entry></feed>
