POWER SYS. ANALYSIS+DESIGN
6th Edition
ISBN: 9780357700907
Author: Glover
Publisher: INTER CENG
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Chapter 6, Problem 6.34P
To determine
Value of voltage V3 using Gauss-Seidel method after second iteration
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microprocers and microcontroler
Design a counter to count-up from 2 to 7 using three of
D Flip Flops
(3) 3-Bit Count up (3 to 5) Using D Flip-Flop:
The State Equation of D Flip-Flop:
Q(t+1)=D(t) => Dn=Qn
Present State
D Flip-Flop
Next State
n
Q2p Q1p Q0p
3
0 1
1
1
Q2n Q1n Q0n D2 D1 D0
0 0 1 0 0
4
1
0
0
1
0
1
1 0
1
5
1 0
1
0
1
1
01
1
D2-Sum(3,4) and don't care X-Sum(0,1,2,6,7)
D1=Sum(5) and don't care X=Sum(0,1,2,6,7)
D0=Sum(4,5) and don't care X=Sum(0,1,2,6,7)
Using K-map to simplify the functions:
D2=Q1+Q0'
D1=Q1'QO
DO=Q1'
XOX
XOX
Q2 10
Q2 01
Q2 1xx
Q0
QO
Qo
D2 Q2
>CK
Q2
D1 Q1
BCD
CK
Q1
DO QF
►CK
Q0
☐ Present State Next State D Flip-Flop
n Q2p Q1p Q0p Q2n Q1n Q0n D2 D1 D0
2 0 1
0
0 1 1 0 1 1
3 0
1
1
1
0
0 1 00
4
1
0
0
1
0
1
1
0
1
5
1
0
1
1
1
0
1 1
0
6
1
1
0
0
1
0
0
1
0
D2
D2=Sum(3,4,5), X=Sum(0,1,7)
D1
Q2
1
Q1
1
0
☑
0
Qo
D2=Q0+Q1'
✗
0
Q1
Consider the following 4×1 multiplexer with inputs:
w0=2, w1=1, w2=x2' and w3=0
And with switches:
S1 x1 and S0=x0
What is the multiplexer output f as a function of x2, x1
and x0?
Chapter 6 Solutions
POWER SYS. ANALYSIS+DESIGN
Ch. 6 - For a set of linear algebraic equations in matrix...Ch. 6 - For an NN square matrix A, in (N1) steps, the...Ch. 6 - Prob. 6.9MCQCh. 6 - Prob. 6.11MCQCh. 6 - Using Gauss elimination, solve the following...Ch. 6 - Prob. 6.9PCh. 6 - Determine the bus admittance matrix (Ybus) for the...Ch. 6 - Prob. 6.34PCh. 6 - Prob. 6.37PCh. 6 - Prob. 6.38P
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- Can you rewrite the solution because it is unclear? AM (+) = 8(1+ 0.5 cos 1000kt +0.5 ros 2000ks) = cos 10000 πt. 8 cos wat + 4 cos wit + 4 cos Wat coswet. -Jet jooort J11000 t = 4 e jqooort jgoort +4e + e +e j 12000rt. 12000 kt + e +e jooxt igoo t te (w) = 8ES(W- 100007) + 8IS (W-10000) USBarrow_forwardCan you rewrite the solution because it is unclear? AM (+) = 8(1+0.5 cos 1000kt +0.5 ros 2000 thts) = cos 10000 πt. 8 cos wat + 4 cos wit + 4 cos Wat coswet. J4000 t j11000rt $14+) = 45 jqooort +4e + e + e j 12000rt. 12000 kt + e +e +e Le jsoort -; goon t te +e Dcw> = 885(W- 100007) + 8 IS (W-10000) - USBarrow_forwardCan you rewrite the solution because it is unclear? Q2 AM ①(+) = 8 (1+0.5 cos 1000πt +0.5 ros 2000kt) $4+) = 45 = *cos 10000 πt. 8 cos wat + 4 cosat + 4 cos Wat coswet. j1000016 +4e -j10000πt j11000Rt j gooort -j 9000 πt + e +e j sooort te +e J11000 t + e te j 12000rt. -J12000 kt + с = 8th S(W- 100007) + 8 IS (W-10000) <&(w) = USB -5-5 -4-5-4 b) Pc 2² = 64 PSB = 42 + 4 2 Pt Pc+ PSB = y = Pe c) Puss = PLSB = = 32 4² = 8 w 32+ 8 = × 100% = 140 (1)³×2×2 31 = 20% x 2 = 3w 302 USB 4.5 5 5.6 6 ms Ac = 4 mi = 0.5 mz Ac = 4 ५ M2 = =0.5arrow_forward
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