A signal's bandwidth is related to its rise time because rising edges usually have the signal's highest frequency components.
The term bandwidth is used and abused in many situations. I recall one meeting where the word was used to refer to: the frequency content of a particular signal, the frequency response of a specific circuit, the speed of our local area network and the human capacity of the organization. “We just don’t have the bandwidth to handle the workload right now.” I don’t think anyone else noticed, but I found it humorous.
Engineers will ask the question “how much bandwidth do I need for that signal?” Typically, the question relates to making sure that the signal can propagate through a component or system and come out the other end without any degradation.
Figure 1 shows a signal passing through a system with finite bandwidth. As shown, there is enough bandwidth in the system to pass the signal unchanged, which is typically the desired outcome. Of course, if the frequency content of the signal exceeds the bandwidth of the system, the output signal can be significantly degraded.
If we are talking basic sinusoidal waveforms, the answer to “how much bandwidth?” may seem obvious: a 1 GHz sine wave needs at least 1 GHz of system bandwidth to be transmitted effectively. Or perhaps not…if we are using the -3 dB point to describe a system bandwidth, the response will be attenuated at that frequency. For example, if I view a 1 GHz sinewave with a 1 GHz oscilloscope, the displayed waveform may have an amplitude that is 70% (-3 dB) of its actual value, depending on how much margin there is in the oscilloscope bandwidth.
Digital signals are even more of a challenge as they include high frequency content associated with the fast, rising edge of the waveform. Eric Bogatin explained the relationship between the rise time of a signal and its bandwidth in Rule of Thumb #1: Bandwidth of a signal from its rise time. Simply put,
where tr is the rise time (10%, 90%) and f3dB is the 3-dB bandwidth.
This equation is exact for a system with a single-pole low-pass response (think “low-pass RC circuit”) and it’s reasonably close for many well-behaved systems [Ref. 1]. I recently found an old Tektronix article [Ref. 2] that discussed vertical amplifier response and rise time. The article showed these two plots of step response, one for the idealized “gaussian response” and the other for a classic lowpass RC circuit (Figure 2).
I found the associated comments interesting because they describe the design tradeoff between rise time and overshoot. (The factor K corresponds to 0.35 from the equation above):
A true gaussian response resolves to tr • bw = 0.32. Because of several factors oscilloscope amplifiers don’t meet the requirement for true gaussian response. “Gaussian” circuits are thus essentially gaussian. Empirical products, arrived through years of research, define these “gaussian” circuits. For a higher gaussian response, tr • bw = 0.35 to 0.45. Higher products indicate least risetime. However, overshoot accompanies the risetime reduction. A product of 0.45 results in 5% overshoot and when tr • bw = 0.35 there is little, if any, overshoot in the step response. Tektronix usually established the product at K = 0.35, sacrificing risetime for minimum overshoot.
K is a figure of merit and is sometimes referred to as the Rise Time Bandwidth Product (RTBP). In recent years, as oscilloscope manufacturers have pushed the upper limits of usable bandwidth, they have sometimes implemented a steeper roll off of the instrument response with a corresponding shift in the RTBP. For example, the Keysight UXR 110 GHz oscilloscope uses a factor of 0.44 to specify the 10%/90% rise time [Ref. 3]. (For a deeper dive into oscilloscope frequency response, see Ref. 4.)
Back to the sine wave
After looking at the digital case, I kept thinking about the simple sinewave. Rise time is most applicable to step functions or square waves but it can be applied to a sine wave (although I admit it may seem a bit contrived). Consider a sine wave with an amplitude of 1 and a peak-to-peak amplitude of 2 (Figure 3).
Let’s calculate the “rise time” as the time it takes for the waveform to go from 90% of the negative peak to 90% of the positive peak. As shown in the figure, this rise time will be t2 – t1. Due to the symmetry of the waveform, we see that
Well isn’t that interesting. The same value of 0.35 pops up as the relationship between the frequency and the rise time of the sinewave. A careful look reveals that it is slightly different: 0.356 versus 0.350 for the single-pole case [Ref. 1]. There seems to be something special about the number 0.35 but not to the level of being a universal constant (e.g., the speed of light).
—Bob Witte is president of Signal Blue LLC, a technology consulting company.