I'm reading Katsuhijo Ogata's book "Discrete Time control Systems", at some point convolution property is used to get the Laplace transform of a function sampled with impulse train. Here I'm using $j=\sqrt{-1}$

Impulse sampled function: $x_m(t)= x(t)\sum_{n=0}^{\infty} \delta (t-nT) \quad $ (1)

Laplace transform: $\mathcal{L} [ x_m(t) ]= X_m (s) \quad $ (2)

Convolution Property: $ X_m(s)= \frac{1}{j 2 \pi}\int_{c-j\infty }^{c+j\infty } X(k) \frac{1}{1-e^{-T(s-k)} }dk \quad $ (3)

I understand a countour like the image below, where $R \rightarrow \infty $ is used to solve (3) enter image description here

The closed integral equals the sum of the two lines integrals ($\Gamma_1$ and $\Gamma_2$). That is:

$\oint{X(k) \frac{1}{1-e^{-T(s-k)} }dk}=\frac{1}{j 2 \pi}\int_{\Gamma_1}{X(k) \frac{1}{1-e^{-T(s-k)} }dk}+\frac{1}{j 2 \pi}\int_{\Gamma_2}{X(k) \frac{1}{1-e^{-T(s-k)} }dk}$

It is assumed that $X(k)$ es a rational function of p-Polynomials with two cases:

1) The denominator is greater in 2 or more orders than the numerator (page 146). That means:

$\lim_{s \rightarrow \infty} sX(s)= 0$

In this case, the integral over the curved part fades as $R \rightarrow \infty$.

$ \lim_{R \rightarrow \infty} \frac{1}{j 2 \pi} \int_{\Gamma_1} {X(k) \frac{1}{1-e^{-T(s-k)} }dk}= 0 \quad$ (4)

So (3) reduces to:

$\oint{X(k) \frac{1}{1-e^{-T(s-k)} }dk} = \int_{\Gamma_2}{X(k) \frac{1}{1-e^{-T(s-k)} }dk}$

2) The denominator is only one order greater than the denominator (page 148), that means:

$\lim_{s \rightarrow \infty} sX(s)= x(0+)$

In this case, the book states that "it can be proved that":

$\lim_{R \rightarrow \infty} \frac{1}{j 2 \pi} \int_{\Gamma_1} {X(k) \frac{1}{1-e^{-T(s-k)} }dk}= -\frac{1}{2} x(0+) \quad $ (5)

So (3) reduces to:

$\oint{X(k) \frac{1}{1-e^{-T(s-k)} }dk} = \int_{\Gamma_2}{X(k) \frac{1}{1-e^{-T(s-k)} }dk} +\frac{1}{2} x(0+) $

My question is about (4) and (5), what is the approach to get to these conclusions?

I'm fully aware that it isn't required to know this to advance with the reading, but i really want to know the reason of those conclusions.

So far I've tried to to define $X(k)$ (for case 2) as :

$X(k)=\frac{\prod_{n=1}^{N} (k-z_n) } {\prod_{n=1}^{N+1} (k-p_n)} $

Set $k=x+jy$ and $y=\pm \sqrt{R^2 - x^2}$ and try to solve the line integral as:

$\int_{c}^{R} X(x+j\sqrt{R^2 - x^2}) \frac{1}{1-e^{-T(s-x-j\sqrt{R^2 - x^2})}}dx + \int_{R}^{c} X(x-j\sqrt{R^2 - x^2}) \frac{1}{1-e^{-T(s-x+j\sqrt{R^2 - x^2})}}dx$

Then trying partial integration with complex functions $g(x)=\frac{1}{1-e^{-T(s-x-j\sqrt{R^2 - x^2})}}$ and $F(x)=X(x+j\sqrt{R^2 - x^2})$ :

$\int_{c}^{R} F(x) g(x) dx= \left( F(x) G(x) \right) \bigg\rvert_{c}^{R} - \int_{c}^{R} f(x) G(x) dx$

Where $f(x)= \frac{d}{dx} F(x)$ and $G(x)=\int g(x)dx$.

Problem with this approach is just getting $G(X)$ is enough to make me to think I'm doing something wrong. Help me with some guide or reference please.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.