System output in cosine notation

Question

If I have a system with the output: $$y(t)=x(t)+0.7x(t-0.4)$$ For $x_2(t)=\cos(56t)$, how can we write $y_2(t)$ in the form $y_2(t)=A\cos(\omega t+B)$ and calculate the unknown values A, B and $\omega$?

My 1st attempt:

I have written $y_2(t)$ as $y_2(t)$ = $x_2(t) + 0.7 x_2(t-0.4) = \cos(56t) + 0.7\cos(56t-22.4)$ and knowing that $e^{j\theta} = \cos(\theta) + j\sin(\theta)$ and that $e^{-j\theta} = \cos(\theta) - j\sin(\theta)$, I wrote $$\cos(56t) + 0.7\cos(56t-22.4) = \frac{e^{j56t}+e^{-j56t}}{2}+\frac{0.7 e^{j56t-22.4}+0.7 e^{-j56t+22.4}}{2}$$

I then considered $z=1+0.7 e^{-22.4j}$ and rewrote the previous expression as $z e^{j56t}+\overline z\ e^{-j56t}=|z|\ e^{j\angle z}\ e^{j56t} +|z|\ e^{-j\angle z}\ e^{-j56t}=|z|\ e^{j(56t+\angle z)}+e^{-j(56t+\angle z)}=|z|\ 2\cos(56t+\angle z)$

My 2nd attempt:

$$y_2(t) = x_2(t) + 0.7 x_2(t-0.4) = \cos(56t) + 0.7\cos(56t-0.4)$$ Use the cosine relation of $\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)$: \begin{align} y_2(t) &= \cos(56t) + 0.7\cos(56t)\cos(0.4)+0.7\sin(56t)\sin(0.4)\\ &=\cos(56t)+0.6447\cos(56t)+0.2726\sin(56t)\\ &=1.6447\cos(56t)+0.2726\sin(56t) \end{align}

But for example $\phi=\text{arctan}(-\frac{b}{a})=-0.16$ and the result I should come to is $\phi = 0.661$ and $c = 0.453$.

Answer 1

EDIT A more elegant and general answer from Matt L.
The same result through brute-force trigonometry follows.

---

• Use the identity $\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)$
• Re-arrange your result, you should then be able to
• use the identity $a\cos(x) + b\sin(x) = c\cos(x+\phi)$

$c = \text{sign}(a)\sqrt{a^2+b^2}$
$\phi = \arctan\left(-\frac{b}{a}\right) \quad\text{if} \ a\neq0$

Results

\begin{align} A &= \sqrt{(1+0.7\cos(56*0.4))^2 + (0.7\sin(56*0.4))^2}\\ B &= \text{arctan}\left(-\frac{0.7\sin(56*0.4)}{1+0.7\cos(56*0.4)}\right)\\ \omega &= 56 \end{align}

w = 56;
a = 1+0.7*cos(w *0.4); 
b = 0.7*sin(w *0.4);

A = sqrt(a^2+b^2); % Matt L: abs(1+0.7*exp(-1i*0.4*w))
B = atan(-b/a); % Matt L: angle(1+0.7*exp(-1i*0.4*w))

t = (0:100);
y = cos(w*t) + 0.7*cos(w*(t-0.4));
y2 = A*cos(w*t + B);

figure(1)
subplot 211
hold on
plot(y)
plot(y2)
legend('y', 'y2')
grid on
subplot 212
plot(y-y2);
title('error')
grid on

Answer 2

A more general and - in my opinion - more straightforward way to arrive at the result is to use the following property of linear time-invariant (LTI) systems:

The response of an LTI system with frequency response $H(j\omega)$ to an input signal $x(t)=\cos(\omega_0t)$ is given by $y(t)=|H(j\omega_0)|\cos\big[\omega_0t+\arg\left\{H(j\omega_0)\right\}\big]$

The given system is clearly LTI, and its frequency response is easily computed as

$$H(j\omega)=1+0.7e^{-j0.4\omega}\tag{1}$$

With $\omega_0=56$, the desired magnitude and phase values are

$$|H(j\omega_0)|=\left|1+0.7e^{-j0.4\cdot 56}\right|=0.45320\tag{2}$$

and

$$\arg\left\{H(j\omega_0)\right\}=\arg\left\{1+0.7e^{-j0.4\cdot 56}\right\}=0.66119 \textrm{ rad}\tag{3}$$

The advantage of this method is that it is equally easy to derive the expression for the output in the more general case when the input-output relation is given by

$$y(t)=\sum_{k=0}^Na_kx(t-t_k)\tag{4}$$