# Analytical Expressions for Frequency Response

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1. Dec 17, 2015

### MarkDECE

Hi,

I am struggling to understand the maths in the text shown below.

(1) How does: r(t) = Acos(wt)+Bsin(wt)
become; sqrt((A^2)+(B^2))cos(wt - arctan(B/A))?

(2) He states that we can represent the input as a phasor and in one of three ways but doesn't say which one.

(3) He then focus on the forced response {R(s)} the laplace transform of the input.

(4) He then then seperates the forced solution from the transient solution by performing a partial faction expansion but I am unsure what is meant by this seperation?

Any help genuinely appreciated.

2. Dec 19, 2015

### deskswirl

1) First write r(t) in rectangular phasor form:
r(t)=Acos(wt)+Bsin(wt)
=Re{A exp(jwt)}-Re{jB exp(jwt)}
=Re{(A-jB) exp(jwt)}
=Re{sqrt(A^2+B^2) exp(-jarctan(B/A)) exp(jwt)}
=sqrt(A^2+B^2)cos(wt-arctan(B/A) //

2) You can see from 1) that I started in trignometric, went to rectangular then used Euler's formula and a property of exponentials to simplify. This can also be done with trig identities but there are extra steps and since you have a linear system this is an easier approach anyways.

3) I believe the author has implicitly assumed a linear time invariant system, G(s), which is stable and hence the transient contribution is negligible in the long term therefore he focuses on the sinusoidal steady-state response only, C_ss(s).

4) partial faction expansion is a method to isolate the "poles" of a rational function. Remeber the location of the poles are growth (or decay) rates of the exponential functions in the time-domain. Incidentally in this example the poles s=+-jw are associated with the output (and input) frequency since sinusoids are the eigenfunctions of LTI systems.