MATERIALS SCIENCE AND ENGINEERING -PACK
10th Edition
ISBN: 9781119721857
Author: Callister
Publisher: WILEY
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Chapter 5, Problem 12QAP
To determine
To prove:
The concentration at position x after diffusion time is also a solution to Equation
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The input reactance of 1/2 dipole with radius of 1/30 is given as shown in figure below,
Assuming the wire of dipole is conductor 5.6*107
S/m, determine at f=1 GHz the
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c- Reflection efficiency when the antenna is
connected to T.L shown in the figure.
Rr
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Avg/4
RL
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[In(l/a) 1.5]
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Find Zeq here. i already had one solution written to me but it's wrong. my main question is. i know that i do the parallel connection first so 2x2 / 2+2 = 1ohm but what i'm asking is since it's an open terminal is R3,2(parallel resistors) in series to R1? or should i first do R3,2 // to ZL and then add R1 in series? PLEASE READ THIS. and solve properly. EXPLAIN WHAT I ASKED PROPERLY. UPVOTE WILL BE GIVEN.
For the given virtual address, indicate the TLB entry accessed, the physical address, and the cache byte
value returned in hex. Indicate whether the TLB misses, whether a page fault occurs, and whether a cache
miss occurs.
If there is a cache miss, enter "-" for "Cache Byte returned". If there is a page fault, enter "-" for “PPN”
and leave parts C and D blank.
Virtual address: 1DDE
A. Virtual address format (one bit per box)
15 14 13 12 11 10 9
8 7 6 5 4 3210
B. Address translation
Parameter
Value
VPN
Ox
TLB Index
Ox
TLB Tag
Ox
TLB Hit? (Y/N)
Page Fault? (Y/N)
PPN
Ox
C. Physical address format (one bit per box)
12 11 10 9 8 7 6 5 4 3 210
D. Physical memory reference
Parameter
Byte offset
Cache Index
Cache Tag
Value
0x
Ox
Ox
Cache Hit? (Y/N)
Cache Byte returned Ox
Chapter 5 Solutions
MATERIALS SCIENCE AND ENGINEERING -PACK
Ch. 5 - Prob. 1QAPCh. 5 - Prob. 2QAPCh. 5 - Prob. 3QAPCh. 5 - Prob. 4QAPCh. 5 - Prob. 5QAPCh. 5 - Prob. 6QAPCh. 5 - Prob. 7QAPCh. 5 - Prob. 8QAPCh. 5 - Prob. 10QAPCh. 5 - Prob. 12QAP
Ch. 5 - Prob. 13QAPCh. 5 - Prob. 14QAPCh. 5 - Prob. 15QAPCh. 5 - Prob. 17QAPCh. 5 - Prob. 18QAPCh. 5 - Prob. 19QAPCh. 5 - Prob. 20QAPCh. 5 - Prob. 21QAPCh. 5 - Prob. 24QAPCh. 5 - Prob. 26QAPCh. 5 - Prob. 27QAPCh. 5 - Prob. 31QAPCh. 5 - Prob. 32QAPCh. 5 - Prob. 34QAPCh. 5 - Prob. 35QAPCh. 5 - Prob. 1DPCh. 5 - Prob. 2DPCh. 5 - Prob. 3DPCh. 5 - Prob. 4DPCh. 5 - Prob. 1SSPCh. 5 - Prob. 2SSPCh. 5 - Prob. 3SSPCh. 5 - Prob. 4SSPCh. 5 - Prob. 1FEQPCh. 5 - Prob. 2FEQP
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- Find Zeq here, ignore the semi circle in the wiring i'm just bad at drawing circuits. ZL=JWL write Zeq in terms of JW and give me the final equation. (basically check the parallel and series combinations and give me the final answer.)Will upvote correct answer. Thanks!arrow_forwardرايدة حل هذا السؤال تكدر ترفعه الي محتاجه حله ضروريarrow_forwardThe following problem concerns the way virtual addresses are translated into physical addresses. Thememory is byte addressable. • Memory accesses are to 1-byte words (not 4-byte words). • Virtual addresses are 16 bits wide. • Physical addresses are 13 bits wide. •The page size is 512 bytes. • The TLB is 8-way set associative with 16 total entries. • The cache is 2-way set associative, with a 4 byte line size and 16 total lines. In the following tables, all numbers are given in hexadecimal. The contents of the TLB, the page table for the first 32 pages, and the cache are as follows: TLB Page Table 2-way Set Associative Cache Index Tag PPN Valid VPN PPN Valid VPN PPN Valid Index Tag Valid Byte 0 Byte 1 Byte 2 Byte 3 Tag Valid Byte 0 Byte 1 Byte 2 Byte 3 0 09 4 1 00 6 1 10 0 1 0 19 1 99 11 23 11 00 0 99 11 23 11 12 2 1 01 5 0 11 5 0 1 15 0 4F 22 EC 11 2F 1 55 59 OB 41 10 0 1 02 3 1 12 2 1 2 1B 1 00 02 04 08 OB 1 01 03 05 07 08 5 03 4 1 13 4 0 3 06 0 84 06 B2 9C 12 0 84 06 B2 9C 05 7 1 04…arrow_forward
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