N Thread Starter Niero Joined Nov 11, 2009 1 Nov 18, 2011 #1 "Deriviation of RMS value of full wave and half wave rectifier signal ???" i do a couple of google search and just can't find an answer waiting for response and any answers will be appreciated thankz in advance.
"Deriviation of RMS value of full wave and half wave rectifier signal ???" i do a couple of google search and just can't find an answer waiting for response and any answers will be appreciated thankz in advance.
t_n_k Joined Mar 6, 2009 5,455 Nov 18, 2011 #2 The traditional math approach .... For the full-wave \(V_{rms}=\sqr{\frac{2}{T}\int_0^{\frac{T}{2}}{(V_msin(\omega t))}^2}dt=\sqr{\frac{2}{T}\int_0^{\frac{T}{2}}{V_m}^2sin^2(\omega t)}dt\) For the half-wave \(V_{rms}=\sqr{\frac{1}{T}\int_0^{\frac{T}{2}}{(V_msin(\omega t))}^2}dt=\sqr{\frac{1}{T}\int_0^{\frac{T}{2}}{V_m}^2sin^2(\omega t)}dt\) \(\omega=\frac{2\pi}{T}\) Use \(sin^2(\omega t)=\frac{1-cos(2\omega t)}{2}\) to do the integration. \(\int \frac{1-cos(2\omega t)}{2}dt=\frac{t}{2}-\frac{sin(2\omega t)}{4\omega}\) You should try to do some of the math work yourself to prove Vrms[full-wave]=Vm/√2 Vrm[half-wave]=Vm/2
The traditional math approach .... For the full-wave \(V_{rms}=\sqr{\frac{2}{T}\int_0^{\frac{T}{2}}{(V_msin(\omega t))}^2}dt=\sqr{\frac{2}{T}\int_0^{\frac{T}{2}}{V_m}^2sin^2(\omega t)}dt\) For the half-wave \(V_{rms}=\sqr{\frac{1}{T}\int_0^{\frac{T}{2}}{(V_msin(\omega t))}^2}dt=\sqr{\frac{1}{T}\int_0^{\frac{T}{2}}{V_m}^2sin^2(\omega t)}dt\) \(\omega=\frac{2\pi}{T}\) Use \(sin^2(\omega t)=\frac{1-cos(2\omega t)}{2}\) to do the integration. \(\int \frac{1-cos(2\omega t)}{2}dt=\frac{t}{2}-\frac{sin(2\omega t)}{4\omega}\) You should try to do some of the math work yourself to prove Vrms[full-wave]=Vm/√2 Vrm[half-wave]=Vm/2