Electrical – how to identify an impulse response in a system

Although you are in frequency domain you still should be able to get all parameters as you were in time domain. They are different domains but they both should represent the same thing. Time domain represent things in terms of amplitude in respect to time. Frequency domain represent things in terms of amplitude AND PHASE in respect to frequency values. Note that you should have both amplitude and phase in frequency domain, since in the time domain the phase can be represented in the same plot by a shift.

One way to represent these things in frequency domain is by dealing with complex numbers. Complex numbers can be viewed as vectors in a 2D space which have a length (as you said) and an angle. The length represents the output/input ratio and the angle represent the phase shift in comparison also to the input.

So, answering your question, you should calculate the H length to find your output/input ratio. To help you, imagine that:

\$e^{jw}=cos(w)+jsin(w)\$

In other words, its a complex number with always length of 1 and angle \$w\$

You can solve this by two methods:

-Vector method:

imagine that number 1 is \$Z=1+0.i\$ which is a vector to the right, with length 1 and angle \$0\$.

Imagine that \$e^{jw}\$ is a vector that I showed right above

Now add them. Then divive vectors 1 by the vector that you've found.

-Cartesian Coordinates:

represent all in terms of \$Z=a+jb\$ and also \$e^{jw}=cos(w)+jsin(w)\$

and imagine that you have:

\$\large Z = \frac{Z_1}{Z_2+Z_3}\$

and then find length of Z by:

\$|Z| = \sqrt{a^2+b^2}\$

When they say:

$$y(n)=x(n)\ast h(n)=\sum_{m=-\infty}^{\infty}x(m)\cdot h(n-m)$$

they are correct, but in the next passage they oversimplify a bit.

Let's do it step by step:

\begin{align*} &\sum_{m=-\infty}^{\infty}x(m)\cdot h(n-m) = \sum_{m=-\infty}^{\infty}u(m)\cdot u(n-m) = \\ &= \sum_{m=0}^{\infty} 1 \cdot u(n-m) = \sum_{m=0}^{\infty} u(n-m) \\ \end{align*}

The function to be summed is \$u(n-m)\$, which is equal to 1 if and only if \$n-m>0 \Leftrightarrow m < n\$.

Since the sum is carried out only for \$m>=0\$, if \$n>=0\$ the result is actually \$n+1\$, but if \$n<0\$ the result is 0, therefore they should have written:

\[ \sum_{m=-\infty}^{\infty}x(m)\cdot h(n-m) = (n+1)u(n) \]

rightaway.

The following image may help:

the red dots are the function values, the green shaded area highlights the value range that will be summed. The image was drawn with \$n = 3\$. You may infer that if \$n<0\$ the green area will disappear, indicating there is nothing to sum.