# How to interpret the notation of a transfer function

Probably a stupid question, but I'm slowly making my way through Julius Smiths' filter book. The maths is slightly defeating me at points.

This function $H(e^{j\omega t})$, why is it a function of a formula?

All my functions at school a decade ago were of a single variable, and you took the thing in the brackets and substituted it in.. it was nice and easy. Here I'm confused.

I'm guessing here it's talking about complex frequency. But why is it written like this? That is just a standard convention in the signal processing literature.

It is written that way because a DTFT is evaluating the z-transform on the unit circle in the complex z-plane: $$H(z)|_{z=e^{i\omega n}} = H(e^{i\omega n})$$

The same convention is followed for CTFT's because a CTFT is evaluating the Laplace transform on the imaginary axis in the complex s-plane: $$H(s)|_{s=i\Omega t} = H(i\Omega t)$$

There is a very nice relation between trigonometric functions and exponential function:

$$e^{i \theta} = \cos(\theta) + i \sin(\theta)$$

$$\cos(\theta) = \dfrac{e^{i\theta}+e^{-i\theta}}{2}$$

$$\sin(\theta) = \dfrac{e^{i\theta}-e^{-i\theta}}{2i}$$

If we use the expansion for $\cos$ function then it becomes trivial for:

$$H(e^{i\omega T}) = (1+e^{-i\omega T})$$

1. Expand your equation in order to get $-i\omega T / 2$ factor in exponent. This will give you a sum of two exponents, multiplied by the other one:

\begin{align} H(e^{i\omega T}) = \left(e^{i\omega T/2}+e^{-i\omega T/2} \right)e^{-i\omega T/2} &= \\e^{i\omega T/2} e^{-i\omega T/2} + e^{-i\omega T/2} e^{-i\omega T/2} &=\\ e^0 + e^{-i\omega T} \end{align}

1. Use the identity mentioned above in order to obtain $\cos$ term, where $\theta = \omega T/2$. Multiply both sides by $2$:

$$2\cos(\omega T/2) = 2\dfrac{e^{i\omega T/2}+e^{-i\omega T/2}}{2}$$

$$2\cos(\omega T/2) =e^{i\omega T/2}+e^{-i\omega T/2}$$

1. Substitute the result:

$$H(e^{i\omega T}) = \left(e^{i\omega T/2}+e^{-i\omega T/2} \right)e^{-i\omega T/2} = 2\cos(\omega T/2) e^{-i\omega T/2}$$

QED.

I interpret the question a little differently and I'm trying answer specically

"This function H(ejωt), why is it a function of a formula?"

This is simply a convention and arguably a little sloppy. The main tool to analyze time discrete systems is the Z-Transform which is the equivalent of the Laplace Transform in the time continuous domain. Hence you'll see a lot $H(z)$ where z is a single (albeit complex) variable. $z$ describes a location in the Z-plane. The most interesting part of the Z-plane is the unit circle, that's where the actual frequencies are that can be observed and measured. The function $z = e^{j \omega}$ with $\omega$ being a real valued variable (normalized frequency) describes exactly the unit circle in the Z plane. So when someone uses $H(e^{j \omega})$ they are saying "I'm mostly interested in what happens on the unit circle". This can be parameterized with a real variable instead of the whole Z plane that's parameterized with a complex variable.

Arguably it would cleaner to write this simply as $H(\omega)$ but $H(e^{j \omega})$ has the advantage that equations stay the same since $z$ and $e^{j \omega}$ are interchangeable.

In practice $H(\omega)$, $H(j\omega)$, $H(e^{j \omega})$, and $H(z)$ all mean basically the same thing and it's more a question of convention and convenience and there is rarely an actual difference.

• Ah - I was really trying to read too much into this. I just did the analysis on a difference equation to find the transfer function, and I'm starting to write down the formula and thinking, I'm writing H(e^jwt) because I know it's the right answer ... but really i'm plugging in the value of omega in here... – JayC Sep 5 '15 at 20:57