9.19 Generate Bode magnitude and phase plots (straight-line approximations) for the following voltage transfer functions. (a) H(
ω) = 4×104(60+j6ω) (4+j2ω)(100+j2ω)(400+j4ω) (b) H(ω) = (1+j0.2ω)2(100+j2ω)2 (jω)3(500+jω) (c) H(ω) = 8×10−2(10+j10ω) jω(16−ω2 +j4ω) (d) H(ω) = 4×104ω2(100−ω2 +j50ω) (5+j5ω)(200+j2ω)3
1 answer:
You might be interested in
Answer:
True
Explanation:
For point in xz plane the stress tensor is given by
where Dx is the direct stress along x ; Dz is direct stress along z ; tzx and txz are the shear stress components
We know that the stress tensor matrix is symmetrical which means that tzx = txz ( obtained by moment equlibrium )
thus we require only 1 independent component of shear stress to define the whole stress tensor at a point in 2D plane
Answer:
The temperature T= 648.07k
Explanation:
T1=input temperature of the first heat engine =1400k
T=output temperature of the first heat engine and input temperature of the second heat engine= unknown
T3=output temperature of the second heat engine=300k
but carnot efficiency of heat engine =
where Th =temperature at which the heat enters the engine
Tl is the temperature of the environment
since both engines have the same thermal capacities <em> </em> therefore
We have now that
multiplying through by T
multiplying through by 300
-
The temperature T= 648.07k