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For an input with limited amplitude, the maximum amplitude gain of a filter can be calculated as the absolute sum of the impulse response Scaling with the inverse of the absolute sum, will guarantee that your max output amplitude will never exceed the max input amplitude In practice, this does NOT work well: It's a very conservative scaling that often ...


3

The answer is yes but one has to specify $B_n$ properly to avoide possible confusions. In case if one uses a pulse compression, the bandwidth through which the receiver collects the noise will normally be $B_n = \beta_c$. Then, the "new" signal-to-noise ratio should be written as: $SNR = \dfrac{P_TG_TG_R\lambda^2\sigma{P_g}}{(4\pi)^3R^4(kT_{sys}\beta_c)} = \...


2

As mentioned in Hilmar's answer, without any extra information, the only scaling that guarantees no increase in signal amplitude is $l_1$-scaling, i.e., scaling by $\sum_n|h[n]|$, where $h[n]$ are the filter coefficients: $$\big|y[n]\big|=\left|\sum_kh[k]x[n-k]\right|\le\max_n\big|x[n]\big|\sum_k\big|h[k]\big|\tag{1}$$ In practice, this type of scaling is ...


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Take the Z Transform of your difference equation and perform algebraic manipulations to find $H(z)=\dfrac{Y(z)}{X(z)}$ as a rational function of $z$. Take the Z Transform of your input $x(nT) = x[n]$ to find $X(z)$. Multiply to get $H(z)X(z) = Y(z)$. Find the inverse Z Transform of $Y(z)$ to find $y[n]$. (This will be the difficult part, as you may need ...


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