LTI and Transfer Functions

Q1: What's the easiest way to find the zero-state solution for a dissipative, LTI system with a sinusoidal input?
A1: By using a transfer function (that makes use of phasors).

Suppose you know that $x(t)=Kcos(\omega t)$ and the linear differential equation with constant coefficients that relates your input to the output that you want to find.
First, we manipulate the differential equation:
$a_n\frac{d^m}{{dt}^m}y(t)+a_{m-1}\frac{d^{m-1}}{{dt}^{m-1}}y(t)+...+a_1\frac{d}{dt}y(t)+a_0y(t)=$
$b_n\frac{d^n}{{dt}^n}x(t)+b_{n-1}\frac{d^{n-1}}{{dt}^{n-1}}x(t)+...+b_1\frac{d}{dt}x(t)+a_0y(t)$
In a more compact form:
$\sum\limits_{i=0}^ma_i\frac{d^i}{dt^i}y(t)=\sum\limits_{k=0}^nb_k\frac{d^k}{dt^k}x(t)$

Now, you can use the fact that $x(t)=Kcos(\omega t)$ by plugging in the corresponding phasor $B$ multiplied by $e^{j\omega t}$. As you'd expect, taking the real part of the phasor gives you back the original sinusoidal input.

After substitution:
$\sum\limits_{i=0}^ma_i\frac{d^i}{dt^i}y(t)=\sum\limits_{k=0}^nb_k\frac{d^k}{dt^k}Be^{j\omega t}$

Even though there are complex coefficients, the equation is still linear. Hence, you can use the guess and check method to determine a solution.

Suppose that the real part of $Ae^{j\omega t}$ corresponds to the output $y(t)$.
After substitution:
$\sum\limits_{i=0}^ma_i\frac{d^i}{dt^i}Ae^{j\omega t}=\sum\limits_{k=0}^nb_k\frac{d^k}{dt^k}Be^{j\omega t}$
$\sum\limits_{i=0}^ma_i(j\omega)^iAe^{j\omega t}=\sum\limits_{k=0}^nb_k(j\omega)^nBe^{j\omega t}$

$A\sum\limits_{i=0}^ma_i(j\omega)^i=B\sum\limits_{k=0}^nb_k(j\omega)^k$

For neatness, you can define to polynomials:
1. $P_D(j\omega)\equiv\sum\limits_{i=0}^ma_i(j\omega)^i$
2. $P_N(j\omega)\equiv\sum\limits_{k=0}^nb_k(j\omega)^k$
$AP_D(j\omega)=BP_N(j\omega)$

We now define the transfer function, which captures the relationship between input and output:
$H(j\omega)\equiv\frac{P_N}{P_D}$
$A=BH(j\omega)$

Converting from the complex phasor corresponding to the output to the real output:
$y(t)=Re\left\{Ae^{j\omega t}\right\}$
$y(t)=Re\left\{BH(j\omega)e^{j\omega t}\right\}$

Q2: I thought the transfer function was ratio of the output to input. Is this new definition in A1 just the same way of saying this?

A2: No, the transfer function as defined in A1 is the ratio of the complex polynomial associated with the output to the complex polynomial associated to the input.

Q3: I noticed that A1 assumes the sinusoid input is a cosine. What about a sine input? Or a combination of the two?
A3: The following table relates sinusoid inputs to output of a dissipative, LTI system:

Input: x(t)	Output: y(t)
$Acos(\omega t) =Re\left\{Ae^{j\omega t}\right\}$	$Re\left\{AH(j\omega)e^{j\omega t}\right\}$
$Bsin(\omega t) =Re\left\{-jBe^{j\omega t}\right\}$	$Re\left\{-jBH(j\omega)e^{j\omega t}\right\}$
$Acos(\omega t)+Bsin(\omega t) =Re\left\{(A-jB)e^{j\omega t}\right\}$	$Re\left\{(A-jB)H(j\omega)e^{j\omega t}\right\}$

Q4: Couldn't I write $Bsin(\omega t)$ as $Im\left\{Be^{j\omega t}\right\}$?

A4: Yes, you could. And then $y(t)$ would be $Im\left\{BH(j\omega)e^{j\omega t}\right\}$.

But for the sake of consistency, we will adopt the convention that uses the real part of the complex number.