# Tag Info

1

Fat32, how do you apply your method to the following Differential Equation?: y''+3y'+2y=x''+x'-2x I tried your method but I seem to get a wrong answer...

0

Everything becomes very clear once you cinsder the following pair of signals and transforms: If $g[n]$ has a Z-transform $G(z)$ then the expanded (zero stuffed) signal $h[n]$ has the Z-transform $H(z) = G(z^N)$, where $$h[n] = \begin{cases} {g[n/N] ~~~,~~~ n = m N \\ 0 ~~~,~~~ \text{otherwise} \tag{1}}\end{cases}$$ Looking at the given Z-transform : $... 0 So finally this is the way I got to plot it right: Gp = ct.tf(100,[1,0,100]) ts = 0.05 Gz =ct.sample_system(Gp, ts) sym.pprint(Gz) t = np.linspace(0,14*ts,num=15) T, yout = ct.impulse_response(Gz, t) yout=yout.flatten() fig, ax = plt.subplots() ax.step(T, yout) ax.yaxis.set_minor_locator(MultipleLocator(0.05)) ax.yaxis.set_major_locator(MultipleLocator(0.... 0 Your confusion should be removed away when you consider the fact that the following signals are the same : $$u[n] - u[n-3] ~ = ~ \delta[n] + \delta[n-1] + \delta[n-2]$$ or generalizing for any integer$M$: $$u[n] - u[n-M] = \delta[n] + \delta[n-1] +...+ \delta[n-M+1] = \sum_{k=0}^{M-1} \delta[n-k]$$ or even further for$K < M$$$u[n-K] - u[n-M] = ... 0 Taking Equation 2.74 from Example 3 and setting M_1=0 gives: h[n] = \frac{1}{M_2+1} \sum_{k=0}^{M_2} \delta[n-k] Now lets take a closer look at this: h[n] is non-zero only for certain values of n. Its a good a starting point as any, so lets look at n=0. We have: h[0] = \frac{1}{M_2+1} \bigg( \delta[0] + \delta[-1] + ... + \delta[-M_2] \... 3 Front/back is actually very hard too, at least for stationary simulation without head tracking. The main reason is simple: Left/Right is done by looking at the differences between the two ear signals, i.e. interaural level differences and interaural time differences. If a source is located to the left, the sound will arrive earlier on the left ear and will ... 4 You're almost there; you just need to connect a few dots. Let your \frac{10}{5 + i 2 \pi f} = G(f). Then g(t) = 10 u(t) e^{-5 t}. Now we get into that exponent part. Your h(t) = g(t - t_0) is correct, but you're applying it incorrectly. You need to apply it to the part that I've labeled as g(t): h(t) = g(t - t_0) = 10 u(t - t_0) e^{-5 (t - t_0)}... 1 For those who prefer an approach using limits instead of integrals, it follows like this: Consider the following definition of the unit-impulse located at the origin t=0 :$$ \delta(t) = \lim_{\Delta \to 0} \delta_{\Delta}(t) \tag{1} $$where the classical function \delta_{\Delta}(t) is defined as$$ \delta_{\Delta}(t) = \begin{cases} {\frac{1}{\... 2 [While commenting on Matt's answer, I tried to find a different path. I somehow failed to do so, but it is written, so] A folk (and false) interpretation of the Dirac$\delta(t)$is that this is would be a function (false, in the classic sense, it cannot be evaluated; it should be understood as an application or operator on other functions, and called ... 5 First of all it seems useful to establish what we mean by an equation like $$\delta(at+b)=\frac{1}{|a|}\delta(t+b/a),\qquad a\neq 0\tag{1}$$ Since the Dirac impulse$\delta(t)$is a distribution, Eq.$(1)\$ only makes sense if interpreted as $$\int_{-\infty}^{\infty}\delta(at+b)\phi(t)dt=\frac{1}{|a|}\int_{-\infty}^{\infty}\delta(t+b/a)\phi(t)dt\tag{2}$$ ...

2

If you’re looking for modeling the amplifier itself, convolution will not provide a complete model for the internal processes. However, convolution is the basis for a number of cab modeling products. I have a line 6 helix that I use frequently. A dry guitar doesn’t sound great. A dry guitar through an amp model sounds bad. A dry guitar through an amp and ...

11

When talking about modeling, there are two things that usually get modeled: 1. the guitar amp, and 2. the speaker cabinet. Only the latter is modeled by an impulse response, which means that the cabinet is simply represented by an LTI system and implemented by convolution. This is of course an approximation but it works fairly well. You can find a lot of ...

8

If you're an EE student, you will have encountered the term LTI System (or you certainly will soon enough!): A system that, no matter the absolute time, outputs, given the same input, the same output; if you scale the input by a factor, the output is scaled by the same factor. Linear, time-invariant, so to speak. LTI systems can be applied to time-domain ...

Top 50 recent answers are included