Discussion in 'Homework Help' started by need_help, Sep 21, 2009.

1. ### need_help Thread Starter New Member

Sep 7, 2009
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0
frequency response

H(ω) = 1/( ( jω+1)*( jω+2) ) A/V

Determine if the system steady output y(t) with the following inputs:

a) f(t)= 4 V DC
b) f(t)=2cos(2t) V
c) F(t)= cos (2t-10°) + 2sin(4t) V

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Hi need_help,

The normal expectation is that you provide some indication of what you have attempted so far - rather than hoping someone will supply the complete solution.

3. ### need_help Thread Starter New Member

Sep 7, 2009
23
0
switch to the Laplace transform representation:

H(s) = 1/[ (s+1) (s+2) ], Laplace transform of system
transfer function.

For the forcing function: f(t) = 4 volt dc input:

the Laplace transfom is L{4} = 4/s,

For f(t) = 2cos2t, => L{ } = 2s/(s²+2²), and,

for f(t) = cos(2t -10) + 2sin4t => L{ } = s[e-(10s)] / (s²+2²)

Y(s) = H(s)U(s) = 4/{ s(s+1)(s+2) }

A/s + B/(s+1) + C/(s+2) =>

A(s+1)(s+2) +Bs(s+2) + Cs(s+1) = 4

collect/arrange terms and equate like powers of s
in the expressions.

As²+Bs²+Cs² = 0s²
3As+Bs+Cs = 0s
2A = 4, => A = 2

Y(s) = 2/s - 4/(s+1) +2/(s+2)

y(t) = L{Y(s)} = 2u(t) - 4u(t-1) +2u(t-2)

Now, for f(t) = 2cos2t => 2s/(s²+2²)

Y(s) = H(s)U(s)

Y(s) = 2s/[ (s+1)(s+2)(s²+2²)=2s/[ (s+1)(s+2)(s-j2)(s+j2)

use partial fraction expansion:

A/s+B/(s+1)+C/(s+2)+D/(s-j2)+E/(s+j2) = 4s

Is there any easier way ???

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Is there any easier way?

The Laplace method gives you a complete solution - both the transient and steady state components of the response are derived. You really only need the steady state responses.

You could apply some "simple' reasoning as follows

1. In the case of DC stimulus, the transfer function H(jw) reduces to a form in which w=0

So H(jw)=1/((1+jw)*(2+jw)) becomes H(0)=1/(1*2)=1/2 [A/V]. The output DC (current?) value is therefore 1/2 of the input DC voltage. Or 2A ..

2. If w=2 then H(j2) = 1/((1+j2)*(2+j2)) = -0.05-j0.15. Now think about what happens if a sinusoidal voltage 2cos(2t) is applied to the transfer function H(j2) - do it in the frequency [not Laplace] domain and then convert back to the time domain.

3. Split F(t) into the two components at w=2 and w=4 - you'll need to also determine H(j4). Determine the output response for each angular frequency. Phase shift due to the (linear) transfer function can be algebraically added to the initial phase values of the forcing function(s). Since the system is linear you can then add the two AC resulting components in 2w and 4w to give the overall response in the time domain.