The input to a matched filter is given by $$$S\left( t \right) = \left\{ {\matrix{ {10\sin \left( {2\pi \times {{10}^6}t} \right),} & {0 < 1 < {{10}^{ - 4}}\sec } \cr {0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,} & {otherwise} \cr } } \right.$$$

The peak amplitude of the filter output is

A modulated signal is given by, $$s\left(t\right)=m_1\left(t\right)\cos\left(2{\mathrm{πf}}_\mathrm c\mathrm t\right)\;+m_2\left(t\right)\sin\left(2{\mathrm{πf}}_\mathrm c\mathrm t\right)$$, where the baseband signal m₁(t) and m₂(t) have bandwidths of 10 kHz and 15 kHz, respectively. The bandwidth of the modulated signal, in kHz, is

The phase margin (in degrees) of a system having the loop transfer function is $$G(s)H(s) = {{2\sqrt 3 } \over {s(s + 1)}}$$

Consider the points s₁ = −3 + j4 and s₂ = −3 − j2 in the s-plane. Then, for a system with the open loop transfer function $$$G\left(s\right)H\left(s\right)=\frac K{\left(s+1\right)^4}$$$