# Finding transfer function by linear system

Here's the system: $$\begin{cases} z(t)=h_1(t)*x(t) \\ w(t)=z(t)+h_3(t)*y(t) \\ y(t)=h_2(t)*w(t) \end{cases}$$ And $$\begin{cases} h_1(t)=\exp(-3t)u(t-t_0) \text{, with t_0 positive} \\ h_2(t)=\exp(-t)u(t) \\ h_3(t)=\exp(-5t)u(t) \end{cases}$$

I want to find transfer function $$\Y(s)/X(s)\$$ by Laplace transform. So I've tried it and this is what I've got so far: \begin{align} Z(s) &= H_1(s)X(s) = X(s)/(s+3) \\ W(s) &= Z(s)+H_3(s)Y(s) = H_2(s)W(s) = \frac{X(s)}{(s+3)} + \frac{Y(s)}{(s+5)} \\ Y(s) &= H_2(s)W(s) = \frac{W(s)}{(s+1)} \end{align} I can't seem to find $$\Y(s)\$$ and $$\W(s)\$$ values, because when I replace $$\Y(s)\$$ with its value in $$\W(s)\$$ function, $$\W(s)\$$ will eventually be cancelled out and so $$\W(s)\$$ will never be found, same goes for $$\Y(s)\$$. Is there something wrong so far or else what can I do to find $$\Y(s)/X(s)\$$?

assuming those "*" symbols are convolution, just algebra, no? \begin{align} Z(s) &= H_1(s)X(s) \\ W(s) &= Z(s)+H_3(s)Y(s) => W(s) = H_1(s)X(s)+H_3(s)Y(s) \\ Y(s) &= H_2(s)W(s) => \frac{Y(s)}{H_2(s)} \end{align}

\begin{align} \frac{Y(s)}{H_2(s)} = H_1(s)X(s)+H_3(s)Y(s) \\ Y(\frac{1}{H(s)}-H_3(s)) = H_1(s)X(s) \\ \frac{Y(s)}{X(s)} => \frac{H_1(s)}{\frac{1}{H_2(s)}-H_3(s)} \end{align}

• Please use latex or text when posting equations Commented Feb 14, 2021 at 0:00
• thank you for updating it, but IMHO it was equally legible before, and adding all the (s) just clutters it. Commented Feb 14, 2021 at 0:52
• Well, the site flagged it as low quality, I could have just deleted the answer, but I thought I'd give it a second chance. To reiterate: images are not preferred. The (s) follow the format of the OP Commented Feb 14, 2021 at 1:07
• @Voltage Spike - again, I appreciate all the time you're putting into making it better, but it's some kid's homework problem. We're all doing this in the spirit of sharing information. Commented Feb 14, 2021 at 4:35
• I'm not really interested in formatting the homework, I'd appreciate it if you wouldn't post images and format the questions that is all, my intention was to show how to use latex Thanks Commented Feb 14, 2021 at 4:35

You have: y(t)=h2(t) * w(t) and w(t)=z(t)+h3(t) * y(t) That implies an equation which has w(t) in both sides; => w(t)=z(t)+h3(t) * (h2(t) * w(t))

In Laplace domain convolutions become multiplications: W = Z + (H3)(H2)W

You solve W from that equation. W = Z/(1-(H3)(H2))

Initially you had Z=X/(s+3) , H3=1/(s+5), H2=1/(s+1). Inserting these gives W as X multiplied by a rational function of s. I guess that's wanted. One multiplicand term more and you have also Y.

The problem: You had never solved the equations. Repeating substitutions lead to nothing.