# 2 identical circuits cascaded/laplace transform and impulse response

1. May 4, 2012

### gethelpelectr

1. The problem statement, all variables and given/known data
Two identical circuits are cascaded, without the second circuit having any loading effect on the first circuit. If the laplace transform of the impulse response of the cascade is 4/(s+1)^2. a)determine the impulse response of each circuit alone as a function of time.
b)if a signal u(t) is applied to the input of the first circuit, determine the output of the second circuit as a function of time

2. Relevant equations

3. The attempt at a solution
a) the impulse response of each circuit is either H(s)=+-(2/s+1) so h(t)=+-2e^-t
b)the laplace transform of the output is (4/(s*(s+1)^2) then do partial fractions and u can easily make inverse laplace and get the answer
These are the soulutions to the problem, however I don't get them. What do they mean by the second circuit does not have any loading effect on the first? And why did they multiply laplace of u(t) with laplace of the cascade to get the laplace of the input? I'm not really understanding the concepts.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 4, 2012

### Staff: Mentor

An individual circuit has an input impedance and an output impedance. When you analyze such a circuit in isolation the assumed source driving its input has zero internal impedance and the circuit's output is an open circuit (infinite impedance). So the resulting transfer function is for the ideal case.

When you cascade two circuits, connecting the output of the first to the input of the next, the first circuit now "sees" the input impedance of the second as a load, while the second circuit "sees" an imperfect source with an impedance. These changed conditions will alter the responses of both circuits.

In practice when designing stages to be cascaded, one generally arranges things so that the input impedance of succeeding stages is much higher than the output impedance of the preceding stages (usually by a factor of 10 or so), so that they are effectively isolating the effects of one from the other and the individual transfer functions are not significantly perturbed.

For this problem you are told to ignore this loading effect, allowing the two transfer functions to stand unaltered.

As for your second question, the problem asks you to evaluate the output of the cascaded circuits in response to a unit step at the input of the first. The overall transfer function is:

$\frac{Vo(s)}{Vi(s)} = H(s)$

so that

$Vo(s) = Vi(s)H(s)$

and Vi(s) is the Laplace Transform of a unit step, which is 1/s.

3. May 4, 2012

### gethelpelectr

You have really helped thanks a lot