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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 ... 1 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 ...