# Frequency Response of 1st Order circuit

I am having a bit of trouble trying to figure this out. The equation I come up with keeps canceling Omega out.

$$Attempt$$

Converting the circuit to the frequency domain the capacitor becomes $\frac{4}{j\omega}$

I then used a current divider to find $\underline y(j\omega)$ = $\underline u(j\omega)$ 6 $\frac{2+\frac{4}{j\omega}}{6+(2+\frac{4}{j\omega})}$

Final Equation: $\frac{12+\frac{24}{j\omega}}{8+\frac{4}{j\omega}}$ = $\frac{3x+6}{2x + 1}$

Simplifying this I get the frequency response to be $\frac{1}{2}$. I am fairly certain that his is not correct.

• $\omega$ doesn't disappear. But your equation is not correct for $y(j\omega)$ – Chu Nov 13 '15 at 7:09
• @Chu Should it be 6 in the numerator instead of what I have? – user3482104 Nov 13 '15 at 7:25
• What does the current divider equation give as an output quantity? – Chu Nov 13 '15 at 7:28
• @Chu Amps. So would it just be that multiplied by the 6 ohms? – user3482104 Nov 13 '15 at 7:48
• Yes. That will give you an expression in $j\omega$ (doesn't cancel out) - of the form: $\frac{a+jb}{c+jd}$ – Chu Nov 13 '15 at 8:01

a) Your start equation is missing a factor 6, which is the value of the resistor.

The frequency response would then be:

$$G(S)=\frac{Y(S)}{U(S)}=\frac{3.(S+2)}{2.S+1}$$ b)When the the frequency is zero the cap is an open circuit and the resulting gain should be 6, which is the value of the the resisor is parallel with the source.

When the frequency approaches infinity, the cap is a short circuit and the gain should be the parallel combination of the two resistor. That would be 12/8, you will reach the same value after calculating the limit with S-->infinity of the equation above.

• We don't give complete answers to what look like homework questions – Chu Nov 13 '15 at 8:22
• @Chu I did no show the whole process, neither the limit calculation. Item b is based on parctical assumptions. Beginner's error, maybe. – Hagah Nov 13 '15 at 8:31

This exercise is actually the determination of an impedance which is a transfer function: $Z(s)=\frac{Y(s)}{U(s)}=R_0\frac{N(s)}{D(s)}$ where Y(s) is the response (the voltage across the current source) and U(s) is the excitation or the stimulus. The raw transfer function is immediate and given by $H_{ref}=(2\Omega+\frac{1}{s0.25F})||6\Omega$. Unfortunately, if you develop this expression, you won't have a low-entropy form in which you can see poles, zeros and gain, if any. To determine this function is a few seconds (without a line of algebra), we will apply the FACTs as described here.

The transfer function of this circuit in dc ($s=0$) is $6\;\Omega$. You obtain this result by opening the capacitor. Then, if you suppress the excitation (the current source is open-circuited), the time constant involving the capacitor is $\tau_2=(6+8)\times 0.25F$. The pole is thus $\omega_p=\frac{1}{\tau_2}$. The zero is obtained with a zeroed response across the current source (a short circuit). The time constant in this mode is $\tau_1=2\times 0.25F$ and $\omega_z=\frac{1}{\tau_1}$. The complete TF is thus $\frac{Y(s)}{I(s)}=6\times\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_z}}$. If you capture this equation and compare it to the reference equation determined before, you obtain perfectly superimposed curves:

You can also rewrite the TF as follows: $Z(s)=6\times\frac{1+0.5s}{1+2s}$ in which the leading term has the dimension of a resistance as it should be.