I have a transfer function, \$H(s) = \frac{-16(s+4)(s-1) }{ (s^2+16)(s+2)}\$. The impulse response is \$-16*(-3/10 * e^{-2t} + \frac{13}{10} cos(4t) + \frac{1}{10}sin(4t))u(t)\$. I am supposed to find an input of the form \$10e^{-at}\$, where \$a\$ is a positive number, so that the response does not have a term of the form \$Ke^{-at}\$. I believe \$a\$ is not necessarily the same number in the input and response. I am sure it is not a trick question. So far I have written out the transfer function \$\frac{(s+4)(s-1)}{(s^2+16)(s+2)(s+a)}\$. I don't think the constants are important for determining the value of \$a\$. My strategy is either the \$s+2\$ term would have to be eliminated or a conjugate \$-Ke^{-at}\$ would have to be added to the response to cancel them out. I have no ideas for eliminating the \$s+2\$ term. I tried to find a number that would make \$(s+2)(s+a)\$ into \$\frac{s}{(s+y)^2+w^2}+ X\frac{w}{(s+y)^2+w^2}\$ in the partial fraction expansion. I tried to use the Heaviside coverup but I didn't have any success. Any ideas on how to proceed would be appreciated.
3 Answers
Let \$\small a=4\$, i.e. use the TF's zero at \$\small s=4\$ to cancel the input signal's \$\frac{1}{s+4}\$ component.
This method of cancellation is not as successful as might first be thought. The danger is, developing a 'dipole' (not an antenna!), which is a pole and zero very close to each other on the s-plane. This is unavoidable since it's almost impossible, in practice, to obtain coincident pole and zero locations. The dipole, thus created, gives rise to a characteristic small amplitude, but persistent 'tail' to the transient response.
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\$\begingroup\$ This is the effect of so-called "blocking zero". \$\endgroup\$ Commented Oct 9, 2018 at 1:05
I would recommend starting with the output signal you have the Laplace domain: \$Y(s) = \frac{160(s+4)(s-1)}{(s^2+16)(s+2)(s+a)}\$.
Then decompose this single fraction into three (or four) partial fractions. One of those partial fractions will have denominator \$s+a\$. Its numerator (called its residue) will be a function of \$a\$. Set its residue to zero and try and solve for \$a\$.
So I misunderstood the question. The alpha in the input and output are strictly the same and so an alpha of 4 will cause the denominator of the input to cancel with the s+4 term in the input.