1 What is the transfer function for any LTI?
1.1 Front
What is the transfer function for any LTI?
In terms of unit impulse response
1.2 Back
\({\displaystyle W(s) = \mathcal{L}(w(t))}\)
where \(w(t)\) is the unit impulse response
2 What is the system function?
2.1 Front
What is the system function?
For LTI
2.2 Back
It’s called transfer function for any LTI
3 What is the transfer function for this system?
3.1 Front
What is the transfer function for this system?
\({\displaystyle p(D)x = f(t)}\), let \(p(D)\)
3.2 Back
\({\displaystyle W(s) = \frac{1}{p(s)}}\)
4 Show the transfer function as the ratio of output to input
4.1 Front
Show the transfer function as the ratio of output to input
\({\displaystyle p(D) x = f(t)}\) with rest IC
4.2 Back
Taking Laplace transform of both sides gives
\({\displaystyle p(s)X(s) = F(s) \implies X(s) = \frac{1}{p(s)} F(s) = W(s) F(s)}\)
Solving \(W(s)\) shows \({\displaystyle W(s) = \frac{X(s)}{F(s)} = \frac{\text{output}}{\text{input}}}\)
5 Find \(p(D)\) so that \(W(s)\) is the transfer function for the system \(p(D)x = f(t)\)
5.1 Front
Find $p(D)$ so that $W(s)$ is the transfer function for the system $p(D)x = f(t)$
\({\displaystyle W(s) = \frac{s}{s^2 + 1}}\)
5.2 Back
The system \(p(D)x = f(t)\) has transfer function \(\frac{1}{p(s)}\). Since \(W(t)\) is not one over a polynomial there is no such polynomial.
Note that \(W(s)\) is the transfer function for the system \(\ddot{x} + x = \dot{y}\), where we consider \(y\) to be the input
6 What is the transfer function for this LC circuit?
6.1 Front
What is the transfer function for this LC circuit?
\({\displaystyle i’’(t) + 4i = v’(t)}\), where \(i\) is the current and \(v\) the voltage
Solve it with 2 methods
6.2 Back
Method 1: We know that the characterization of the transfer function as the ratio output/input is \(W(s) = I(s)/V(s)\)
Assuming rest IC, we have \(\mathcal{L}(v’) = sV(s)\), so applying the Laplace transform
\({\displaystyle (s^2 + 4) I(s) = sV(s) \implies W(s) = \frac{I(s)}{V(s)} = \frac{s}{s^2 + 4}}\)
We can not write with the form of \(1/p(s)\)
Method 2: We can deduce the \(\mathcal{L}(\delta’(t)) = s \mathcal{L}(\delta(t)) - \delta(0^-) = s\), and solve as usual
7 What is the Laplace transform of this expression?
7.1 Front
What is the Laplace transform of this expression?
\({\displaystyle \mathcal{L}(\delta’(t))}\)
7.2 Back
\({\displaystyle s \mathcal{L}(\delta(t)) - \delta(0^-) = s}\)
8 What is the Green’s Formula in Frequency?
8.1 Front
What is the Green’s Formula in Frequency?
8.2 Back
Let \(p(D)x = f(t)\), with rest IC be the system. Applying Laplace transform for both sides we get
\({\displaystyle p(s)X(s) = F(s) \implies X(s) = \frac{1}{p(s)} F(s) = W(s)F(s)}\)
where \(W(s)\) is the transfer function
9 How can we write convolution of 2 function as a multiplication?
9.1 Front
How can we write convolution of 2 function as a multiplication?
\(w * f\)
9.2 Back
Applying Laplace transform
\({\displaystyle \mathcal{L}(f * g) = W(s)F(s)}\)
Convolution is a type of multiplication, because viewed from the frequency side it is multiplication
10 Show that this equation is true
10.1 Front
Show that this equation is true
\({\displaystyle \mathcal{L}(f * g) = F(s)G(s)}\)
10.2 Back
\begin{align*} \mathcal{L}(f * g) &= \int_0^{\infty} (f * g)e^{-st} \dd{t} \\\ &= \int_0^{\infty} \int_0^t f(t - u)g(u) e^{-st} \dd{u} \dd{t} \\\ &= \int_0^{\infty} \int_u^{\infty} f(t - u)g(u) e^{-st} \dd{t} \dd{u} \qq{Changing order of integration} \\\ &= \int_0^{\infty} \int_0^{\infty} f(v) g(u)e^{-s(v+u)} \dd{v} \dd{u} \qq{Substituting} v=t-u, \dd{v} = \dd{t} \\\ &= \int_0^{\infty} f(v)e^{-sv} \dd{v} \int_0^{\infty} g(u) e^{-su} \dd{u}\\\ &= F(s)G(s) \end{align*}
11 What is the Laplace of this expression?
11.1 Front
What is the Laplace of this expression?
\({\displaystyle \mathcal{L}\biggl(\int_{0^-}^{t^+} f(\tau) \dd{t} \biggr)}\)
11.2 Back
\({\displaystyle \mathcal{L}\biggl(\int_{0^-}^{t^+} f(\tau) \dd{t} \biggr) = \mathcal{L}(f * 1) = F(s) \mathcal{L}1 = \frac{F(s)}{s}}\)
12 What does mean this diagram?
12.1 Front
What does mean this diagram?
12.2 Back
\(\text{input } F(s) \leadsto \text{ output } X(s) = W(s)F(s)\)
13 What is a cascaded system?
13.1 Front
What is a cascaded system?
13.2 Back
There are 2 system where the output of the first, it’ll be the input of the second system
\(p_1(D)x = f\), \(p_2(D)y = x\), with rest IC
The input to the cascade is \(f\) and the output is \(y\)
14 What is the solution of this system?
14.1 Front
What is the solution of this system?
\(p_1(D)x = f\), \(p_2(D)y = x\), with rest IC
14.2 Back
Let \({\displaystyle W_1(s) = \frac{1}{p_1(s)}}\) and \({\displaystyle W_2(s) = \frac{1}{p_2(s)}}\) be the transfer function for the two differential equations
\({\displaystyle X(s) = W_1(s)F(s) \qq{and} Y(s) = W_2(s)X(s)}\), so \({\displaystyle Y(s) = W_2(s)W_1(s)F(s)}\)
The transfer function for the cascade is
\({\displaystyle \text{output / input } = Y(s)/F(s) = W_2(s) W_1(s)}\)
15 Represent with a diagram this cascade system
15.1 Front
Represent with a diagram this cascade system
\(p_1(D)x = f\), \(p_2(D)y = x\), with rest IC
15.2 Back
16 What is the solution for this system diagram?
16.1 Front
What is the solution for this system diagram?
Explain the diagram too
16.2 Back
This is a parallel system. The plus sign in the circle indicates the 2 signals coming into the junction should be added. The split near the start indicates the input \(F(s)\) go into each system
\({\displaystyle X_1(s) = W_1(s) F(s)}\), \({\displaystyle X_2(s) = W_2(s) F(s)}\), \({\displaystyle Y = X_1 + X_2}\)
So the solution
\({\displaystyle Y(s) = W_1 F + W_2 F = (W_1 + W_2) F}\)
The transfer function of the system is \(W_1 + W_2\)
17 How could we create a feedback loop for this system?
17.1 Front
How could we create a feedback loop for this system?
Name both systems
17.2 Back
The original system is known as the open loop system and the corresponding system with feedback is known as the closed loop system
18 What does mean the symbol \(\cross g\) in this system diagram?
18.1 Front
What does mean the symbol $\cross g$ in this system diagram?
18.2 Back
The symbol \(\cross g\) means the input is scaled by \(g\), that is apply a gain of \(g\) to the input.
19 What does mean the symbol \(\sum\) in this system diagram?
19.1 Front
What does mean the symbol $\sum$ in this system diagram?
19.2 Back
The symbol \(\sum\) means the two inputs are combined; the plus and minus signs indicate to add or subtract the corresponding input
20 What is the transfer function for this system?
20.1 Front
What is the transfer function for this system?
20.2 Back
\({\displaystyle V = F - gY}\), \({\displaystyle Y = M \cdot V}\)
That implies
\({\displaystyle Y = W (F - gY) = \frac{W}{1 + gW} \cdot F}\)
So, the transfer function is the ratio output/input
\({\displaystyle \frac{Y}{F} = \frac{W}{1 + gW}}\)
This formula is one case of what is often called Black’s formula
21 What is the partial fraction of this rational function?
21.1 Front
What is the partial fraction of this rational function?
\({\displaystyle \frac{1}{s^4 - 1}}\)
21.2 Back
\({\displaystyle \frac{1}{(s^2 + 1)(s - 1)(s + 1)} = - \frac{1}{2(s^2 + 1)} - \frac{1}{4(s+1)} + \frac{1}{4(s - 1)}}\)