New answers tagged

2

You're right, the ROC of the Laplace transform of a two-sided signal is a strip in the complex plane. In your case, the imaginary axis is inside the ROC, and the ROC is limited by the poles in the right and left half-planes. If the ROC were the right half-plane, the signal would be right-sided, which is clearly not the case.


1

To the extent you can factor the transfer function into individual integrator sections of the general form $\frac{1}{s}$ you can make this substitution, which is an approximation of the Matched-$z$ Transform where you substitute every $s$ for $s=\frac{\ln(z)}{T}$. (map from $s$ to $z$ using $z =e^{sT}$). This results in first order forms given by $H_\mathrm{...


2

Let's go through a few ways to solve this: Fourier transform: an ideal integrator is an LTI system, so its response to a sinusoidal input signal is a sinusoid with the amplitude and phase changed according to the frequency response evaluated at the input frequency (if it exists). For the ideal integrator we have $$H(\omega)=\pi\delta(\omega)+\frac{1}{j\...


2

Another way to see how the forward Euler method approximates a continuous-time system is by considering the "ideal" mapping of the $s$-plane to the $z$-plane (why?): $$z=e^{sT}\tag{1}$$ For frequencies that are much smaller than the sampling frequency (i.e., $|s|T\ll 1$) we can approximate $e^{sT}$ by its first order Taylor series: $$z\approx 1+sT\tag{2}$$...


0

Multiplication with $s$ in the Laplace transform domain equals differentiation in the time domain. In the discrete-time domain we can approximate differentiation by the equation $$y[n]=\frac{x[n+1]-x[n]}{T}\tag{1}$$ where $T$ is the sampling interval. In the Z-transform domain, Eq. $(1)$ becomes $$Y(z)=X(z)\frac{z-1}{T}\tag{2}$$ I.e., the transfer ...


0

Two conjugate poles: $(z-(0.51+0.7i))(z-(0.51-0.7i)) = z^{2} - 1.02z +0.76$ Two conjugate zeros: $(z-(0.57+0.78i))(z-(0.57-0.78i)) = z^{2} - 1.14z +0.94$


0

Why the fear of complex numbers? Just do a distinction of cases and generate solutions for the real and the complex case. $b-a$ in the denominator will always be real, as the imaginary parts will cancel out each other, so complex numbers will only be in the exponents, wich will lead to some nice sinusoidal functions, which is expected, as this is an ...


2

Your analog transfer function looks OK. For the sake of clarity - and to reduce the chance of making errors - I'd just rewrite it as $$H_a(s)=G\cdot\frac{s^2+as + b}{s^2+cs + d}\tag{1}$$ with $$\begin{align}G&=\frac{2R_g}{R_d+2R_g}\\a&=\frac{R_d}{L}\\b&=\frac{1}{LC}\\c&=G\left(a+\frac{1}{2R_gC}\right)\\d&=G\cdot b\frac{}{}\end{align}$$ ...


Top 50 recent answers are included