2
$\begingroup$

There is a system which shifts frequencies of input by -Fc such that: enter image description here

Y(S) = X(S).H(S)

But X(S) has value zero from 0 to Fc. I am confused on how the product of X(S) and H(S) becomes a positive value in Y(S) in that frequency range, for any H(S) ? How transfer function of the system H(S) will look like in the frequency domain ?

$\endgroup$
3
$\begingroup$

While the answers that point out that a system needs to be LTI to have a transfer function is correct, there isn't a lack of trying.

Tymerski, Richard. "Application of the time-varying transfer function for exact small-signal analysis." IEEE Transactions on Power Electronics 9.2 (1994): 196-205.

Kamen, Edward W., Pramod P. Khargonekar, and K. R. Poolla. "A transfer-function approach to linear time-varying discrete-time systems." SIAM journal on control and optimization 23.4 (1985): 550-565.

these are only a few Google hits. There is more.

$\endgroup$
2
$\begingroup$

The system you're looking for cannot be described by a transfer function because it is a time-varying system. Only linear time-invariant (LTI) system can be fully characterized by a transfer function. However, there is no LTI system that can shift frequencies. The output of a (stable) LTI system can only have frequency components that are already present in the input signal.

Frequency shifts are usually achieved by modulation, i.e., by multiplying the input signal with a sinusoid or with a complex exponential. Such a system is linear but not time-invariant.

$\endgroup$
1
$\begingroup$

The frequency shifting system is not LTI, and therefore neither its impulse response nor its frequency response exists. A continuous-time frequency shifting system can be described by the following I/O relationship:

$$ y(t) = \mathcal{T} \{x(t) \} = e^{j\omega_0 t } x(t) $$ which is clearly not time-invariant and hence not LTI.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.