# Is it possible to prove that dither VS PWM can result the same output?

Assume that we have a SISO transfer function model $$G(s)$$ and we want to look how the output looks like when we have a input signal $$u(t)$$ that looks like this:

$$u_1(t) = \left\{\begin{matrix} 50 + 5\sin(2\pi10 t) & u(t) > 0 \\ 0 & u(t) \leq 0 \end{matrix}\right.$$

$$u_2(t) = \left\{\begin{matrix} A_2\sin(2\pi\omega_2 t) & u(t) > 0 \\ 0 & u(t) \leq 0 \end{matrix}\right.$$

Is it possible that I can prove that $$G(s)U_1(s) = G(s)U_2(s)$$ or $$G(s)U_1(s) \sim G(s)U_2(s)$$ by changing $$\omega_2$$ and $$A_2$$?

$$U_i(s), i = 1, 2$$ is the frequency domain signal on the view of $$u_i(t), i = 1, 2$$.

• um, what's $u(t)$ here, then. Is it deterministic? Is it linked to either of $u_1$ or $u_2$? Without very specific restrictions on $u(t)$, your desired statements are plain wrong; without any restricitions, it's trivially true (set $u(t) = -1$ for all $t$). – Marcus Müller Jun 10 '20 at 11:12
• @MarcusMüller $u$ is the input signal to the system $G(s)$. Control engineering. – Daniel Mårtensson Jun 10 '20 at 11:14
• well, point is that if you can't define what $u(t)$ is, your statements are plain wrong – Marcus Müller Jun 10 '20 at 11:15
• @MarcusMüller Input signal? – Daniel Mårtensson Jun 10 '20 at 11:16
• Daniel, sorry misunderstanding with "you can't define $u(t)$" I didn't mean "you can't explain $u(t)$" but "you can't set $u(t)$ so that you end up in a corner case where there might be solutions". – Marcus Müller Jun 10 '20 at 12:39

$$U_1(\omega)$$ contains two frequency peaks (10Hz and 0Hz) and $$U_2(\omega)$$ only one at $$\omega _2$$. $$G(\omega)$$ can change the relative height of the peaks but the overall but not create new ones. $$A_2$$ and $$U_2(\omega)$$ can vary the height and position of the peak but not the number of peaks.
• So there is no way I can get a similar output from $G(s)$ if I use both the output signals where $u_2$ is tuned with $\omega_2$? – Daniel Mårtensson Jun 10 '20 at 12:14
• @DanielMårtensson exactly what the answer says. (and what I've been saying in the comments). ("similar" is a problematic term: it's not really well-defined. Your $\sim$ means "proportional", and yes, we can definitely preclude that). – Marcus Müller Jun 10 '20 at 12:38