Bode plot

Amplitude decibels is usually done using ${\displaystyle {\text{dB}}=20\log _{10}(X)}$ to define decibels. Given a transfer function in the form

${\displaystyle H(s)=A\prod {\frac {(s-x_{n})^{a_{n}}}{(s-y_{n})^{b_{n}}}},}$

where ${\displaystyle x_{n}}$ and ${\displaystyle y_{n}}$ are constants, ${\displaystyle s=\mathrm {j} \omega }$, ${\displaystyle a_{n},b_{n}>0}$, and ${\displaystyle H}$ is the transfer function:

• At every value of s where ${\displaystyle \omega =x_{n}}$ (a zero), increase the slope of the line by ${\displaystyle 20a_{n}\ {\text{dB}}}$ per decade.
• At every value of s where ${\displaystyle \omega =y_{n}}$ (a pole), decrease the slope of the line by ${\displaystyle 20b_{n}\ {\text{dB}}}$ per decade.
• The initial value of the graph depends on the boundaries. The initial point is found by putting the initial angular frequency ${\displaystyle \omega }$ into the function and finding ${\displaystyle |H(\mathrm {j} \omega )|}$.
• The initial slope of the function at the initial value depends on the number and order of zeros and poles that are at values below the initial value, and is found using the first two rules.

To handle irreducible 2nd-order polynomials, ${\displaystyle ax^{2}+bx+c}$ can, in many cases, be approximated as ${\displaystyle ({\sqrt {a}}x+{\sqrt {c}})^{2}}$.

Note that zeros and poles happen when ${\displaystyle \omega }$ is equal to a certain ${\displaystyle x_{n}}$ or ${\displaystyle y_{n}}$. This is because the function in question is the magnitude of ${\displaystyle H(\mathrm {j} \omega )}$, and since it is a complex function, ${\displaystyle |H(\mathrm {j} \omega )|={\sqrt {H\cdot H^{*}}}}$. Thus at any place where there is a zero or pole involving the term ${\displaystyle (s+x_{n})}$, the magnitude of that term is ${\displaystyle {\sqrt {(x_{n}+\mathrm {j} \omega )(x_{n}-\mathrm {j} \omega )}}={\sqrt {x_{n}^{2}+\omega ^{2}}}}$.

To correct a straight-line amplitude plot:

• At every zero, put a point ${\displaystyle 3a_{n}\ {\text{dB}}}$ above the line.
• At every pole, put a point ${\displaystyle 3b_{n}\ {\text{dB}}}$ below the line.
• Draw a smooth curve through those points using the straight lines as asymptotes (lines which the curve approaches).

Note that this correction method does not incorporate how to handle complex values of ${\displaystyle x_{n}}$ or ${\displaystyle y_{n}}$. In the case of an irreducible polynomial, the best way to correct the plot is to actually calculate the magnitude of the transfer function at the pole or zero corresponding to the irreducible polynomial, and put that dot over or under the line at that pole or zero.

Given a transfer function in the same form as above,

${\displaystyle H(s)=A\prod {\frac {(s-x_{n})^{a_{n}}}{(s-y_{n})^{b_{n}}}},}$

the idea is to draw separate plots for each pole and zero, then add them up. The actual phase curve is given by

${\displaystyle \varphi (s)=-\arctan {\frac {\operatorname {Im} [H(s)]}{\operatorname {Re} [H(s)]}}.}$

To draw the phase plot, for each pole and zero:

• If ${\displaystyle A}$ is positive, start line (with zero slope) at 0°.
• If ${\displaystyle A}$ is negative, start line (with zero slope) at −180°.
• If the sum of the number of unstable zeros and poles is odd, add 180° to that basis.
• At every ${\displaystyle \omega =|x_{n}|}$ (for stable zeros ${\displaystyle -\operatorname {Re} (z)<0}$), increase the slope by ${\displaystyle 45a_{n}}$ degrees per decade, beginning one decade before ${\displaystyle \omega =|x_{n}|}$ (e.g., ${\displaystyle |x_{n}|/10}$).
• At every ${\displaystyle \omega =|y_{n}|}$ (for stable poles ${\displaystyle -\operatorname {Re} (p)<0}$), decrease the slope by ${\displaystyle 45b_{n}}$ degrees per decade, beginning one decade before ${\displaystyle \omega =|y_{n}|}$ (e.g., ${\displaystyle |y_{n}|/10}$).
• "Unstable" (right half-plane) poles and zeros (${\displaystyle \operatorname {Re} (s)>0}$) have opposite behavior.
• Flatten the slope again when the phase has changed by ${\displaystyle 90a_{n}}$ degrees (for a zero) or ${\displaystyle 90b_{n}}$ degrees (for a pole).
• After plotting one line for each pole or zero, add the lines together to obtain the final phase plot; that is, the final phase plot is the superposition of each earlier phase plot.

The magnitude (in decibels) of the transfer function above (normalized and converted to angular-frequency form), given by the decibel gain expression ${\displaystyle A_{\text{vdB}}}$:

${\displaystyle {\begin{aligned}A_{\text{vdB}}&=20\log |H_{\text{lp}}(\mathrm {j} \omega )|\\&=20\log {\frac {1}{\left|1+\mathrm {j} {\frac {\omega }{\omega _{\text{c}}}}\right|}}\\&=-20\log \left|1+\mathrm {j} {\frac {\omega }{\omega _{\text{c}}}}\right|\\&=-10\log \left(1+{\frac {\omega ^{2}}{\omega _{\text{c}}^{2}}}\right).\end{aligned}}}$

Then plotted versus input frequency ${\displaystyle \omega }$ on a logarithmic scale, can be approximated by two lines, forming the asymptotic (approximate) magnitude Bode plot of the transfer function:

• The first line for angular frequencies below ${\displaystyle \omega _{\text{c}}}$ is a horizontal line at 0 dB, since at low frequencies the ${\displaystyle \omega /\omega _{\text{c}}}$ term is small and can be neglected, making the decibel gain equation above equal to zero.
• The second line for angular frequencies above ${\displaystyle \omega _{\text{c}}}$ is a line with a slope of −20 dB per decade, since at high frequencies the ${\displaystyle \omega /\omega _{\text{c}}}$ term dominates, and the decibel gain expression above simplifies to ${\displaystyle -20\log(\omega /\omega _{\text{c}})}$, which is a straight line with a slope of −20 dB per decade.

These two lines meet at the corner frequency ${\displaystyle \omega _{\text{c}}}$. From the plot, it can be seen that for frequencies well below the corner frequency, the circuit has an attenuation of 0 dB, corresponding to a unity pass-band gain, i.e. the amplitude of the filter output equals the amplitude of the input. Frequencies above the corner frequency are attenuated – the higher the frequency, the higher the attenuation.

The phase Bode plot is obtained by plotting the phase angle of the transfer function given by

${\displaystyle \arg H_{\text{lp}}(\mathrm {j} \omega )=-\tan ^{-1}{\frac {\omega }{\omega _{\text{c}}}}}$

versus ${\displaystyle \omega }$, where ${\displaystyle \omega }$ and ${\displaystyle \omega _{\text{c}}}$ are the input and cutoff angular frequencies respectively. For input frequencies much lower than corner, the ratio ${\displaystyle \omega /\omega _{\text{c}}}$ is small, and therefore the phase angle is close to zero. As the ratio increases, the absolute value of the phase increases and becomes −45° when ${\displaystyle \omega =\omega _{\text{c}}}$. As the ratio increases for input frequencies much greater than the corner frequency, the phase angle asymptotically approaches −90°. The frequency scale for the phase plot is logarithmic.

The horizontal frequency axis, in both the magnitude and phase plots, can be replaced by the normalized (nondimensional) frequency ratio ${\displaystyle \omega /\omega _{\text{c}}}$. In such a case the plot is said to be normalized, and units of the frequencies are no longer used, since all input frequencies are now expressed as multiples of the cutoff frequency ${\displaystyle \omega _{\text{c}}}$.

Figures 6 and 7 illustrate the gain behavior and terminology. For a three-pole amplifier, Figure 6 compares the Bode plot for the gain without feedback (the open-loop gain) A_OL with the gain with feedback A_FB (the closed-loop gain). See negative feedback amplifier for more detail.

In this example, A_OL = 100 dB at low frequencies, and 1 / β = 58 dB. At low frequencies, A_FB ≈ 58 dB as well.

Because the open-loop gain A_OL is plotted and not the product β A_OL, the condition A_OL = 1 / β decides f_0 dB. The feedback gain at low frequencies and for large A_OL is A_FB ≈ 1 / β (look at the formula for the feedback gain at the beginning of this section for the case of large gain A_OL), so an equivalent way to find f_0 dB is to look where the feedback gain intersects the open-loop gain. (Frequency f_0 dB is needed later to find the phase margin.)

Near this crossover of the two gains at f_0 dB, the Barkhausen criteria are almost satisfied in this example, and the feedback amplifier exhibits a massive peak in gain (it would be infinity if β A_OL = −1). Beyond the unity gain frequency f_0 dB, the open-loop gain is sufficiently small that A_FB ≈ A_OL (examine the formula at the beginning of this section for the case of small A_OL).

Figure 7 shows the corresponding phase comparison: the phase of the feedback amplifier is nearly zero out to the frequency f₁₈₀ where the open-loop gain has a phase of −180°. In this vicinity, the phase of the feedback amplifier plunges abruptly downward to become almost the same as the phase of the open-loop amplifier. (Recall, A_FB ≈ A_OL for small A_OL.)

Comparing the labeled points in Figure 6 and Figure 7, it is seen that the unity gain frequency f_0 dB and the phase-flip frequency f₁₈₀ are very nearly equal in this amplifier, f₁₈₀ ≈ f_0 dB ≈ 3.332 kHz, which means the gain margin and phase margin are nearly zero. The amplifier is borderline stable.

Figures 8 and 9 illustrate the gain margin and phase margin for a different amount of feedback β. The feedback factor is chosen smaller than in Figure 6 or 7, moving the condition | β A_OL | = 1 to lower frequency. In this example, 1 / β = 77 dB, and at low frequencies A_FB ≈ 77 dB as well.

Figure 8 shows the gain plot. From Figure 8, the intersection of 1 / β and A_OL occurs at f_0 dB = 1 kHz. Notice that the peak in the gain A_FB near f_0 dB is almost gone.[note 2][9]

Figure 9 is the phase plot. Using the value of f_0 dB = 1 kHz found above from the magnitude plot of Figure 8, the open-loop phase at f_0 dB is −135°, which is a phase margin of 45° above −180°.

Using Figure 9, for a phase of −180° the value of f₁₈₀ = 3.332 kHz (the same result as found earlier, of course[note 3]). The open-loop gain from Figure 8 at f₁₈₀ is 58 dB, and 1 / β = 77 dB, so the gain margin is 19 dB.

Stability is not the sole criterion for amplifier response, and in many applications a more stringent demand than stability is good step response. As a rule of thumb, good step response requires a phase margin of at least 45°, and often a margin of over 70° is advocated, particularly where component variation due to manufacturing tolerances is an issue.[9] See also the discussion of phase margin in the step response article.

• Examples
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  Figure 6: Gain of feedback amplifier A_FB in dB and corresponding open-loop amplifier A_OL. Parameter 1/β = 58 dB, and at low frequencies A_FB ≈ 58 dB as well. The gain margin in this amplifier is nearly zero because | βA_OL| = 1 occurs at almost f = f_180°.
• 
  Figure 7: Phase of feedback amplifier °A_FB in degrees and corresponding open-loop amplifier °A_OL. The phase margin in this amplifier is nearly zero because the phase-flip occurs at almost the unity gain frequency f = f_0 dB where | βA_OL| = 1.
• 
  Figure 8: Gain of feedback amplifier A_FB in dB and corresponding open-loop amplifier A_OL. In this example, 1 / β = 77 dB. The gain margin in this amplifier is 19 dB.
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  Figure 9: Phase of feedback amplifier A_FB in degrees and corresponding open-loop amplifier A_OL. The phase margin in this amplifier is 45°.