Bandwidth Versus Video Resolution
Abstract: This article discusses the key relationship between video resolution and the required bandwidth to accurately process and display that video signal. The equations and table address standard definition, NTSC and PAL, as well as high definition DTV standards. Computer display formats are also covered. Slew rate requirements are also discussed.
First, we will clarify a common confusion between visual resolution and format resolution. Visual resolution relates to the amount of detail that can be seen and is specified in terms of TV lines (abbreviated TVL), whereas format resolution pertains only to the specified format of the signal. For example, an XGA format computer signal has a format resolution of 1024 horizontal pixels and 768 vertical pixels and a maximum visual resolution of 538 TVL. If this signal is bandlimited to 20MHz, its visual resolution will drop down to 377 TVL and it will not be possible to view all of the detail that is present in this format. Another example is a standard NTSC video signal. The typical horizontal resolution is about 330 TVL. If this signal, for example, is bandlimited to 3MHz instead of the maximum of 4.2MHz, it will have a visual resolution of only 240 TVL. This illustrates the importance of paying close attention to the bandwidth of all devices in the signal path of video signals.
The highest frequency contained in a video signal, and therefore in the signal bandwidth, is a function of the scanning system—meaning, the number of scanning lines, the refresh rate, and so forth. It can be calculated with the following equation:
BW_{S} = 1/2 [(K × AR × (V_{LT})² × F_{R}) × (K_{H} / K_{V})] EQ 1Where:
BW_{S} = Total signal bandwidth
K = Kell factor
AR = Aspect ratio (the width of the display divided by the height of the display)
V_{LT} = Total number of vertical scan lines
F_{R} = Frame rate or refresh rate
K_{H} = Ratio of total horizontal pixels to active pixels
K_{V} = Ratio of total vertical lines to active lines
The 1/2 factor comes from the highest frequency component in a video signal occurring when alternating black and white vertical lines with a width of one pixel are displayed on the screen. Because it takes two lines to form a complete cycle, the highest frequency is onehalf the pixel rate.
The Kell factor represents the effect of reduced visual resolution primarily due to the linescanning structures. Visual information is lost due to the probability that some of the video information will be displayed during the retrace instead of the active portion of the scan line. Even though it may seem like half the information would be lost because there are equal number of scan and retrace lines, empirically it has been shown that about 30% is lost to this effect, yielding a Kell factor of about 0.7.
The frame rate is the rate at which each complete set of scan lines is displayed. Because a set of scan lines makes a complete picture, this can be thought of as the pictureupdate rate. Most television signals are in an interlaced format. This is where each picture (or frame) is divided into two fields, each with half of the vertical scan lines. This doesn't affect the calculation as long as the actual frame rate is used in the equation. Just remember that the frame rate is equal to half the field rate.
K_{V} represents the ratio of the total number of vertical lines divided by the number of active lines. The difference between these is the vertical blanking lines. Similarly, the K_{H} term is the ratio of the total horizontal pixels to active pixels. If the K_{H} and K_{V} values in the above equations are not known, they can be approximated or inferred from the values in the table below.
Visual Resolution
Visual resolution is a measure of the smallest detail that can be seen. TV lines, and therefore resolution, are defined as the number of alternating lines that can be discerned in a width of the screen equal to one picture height. Stated another way, it is the number of visible horizontal pixels divided by the aspect ratio, which is 4:3 for standard TV and 16:9 for digital TV.The visual resolution can be calculated from the signal bandwidth (BW_{S}) by using the following equation:
TVL = (2 t_{HA} BW_{S})/AR EQ 2Where:
TVL = Horizontal resolution specified in TV lines
t_{HA} = Active horizontal period
BW_{S} = Signal bandwidth
AR = Aspect ratio (the width of the display divided by the height of the display)
The active horizontal period is the time it takes to display the active picture portion of one scan line. It is the total time for one scan line minus the retrace time. It can also be expressed as the total horizontal time divided by the K_{H} factor, defined earlier.
The following table uses the equations defined above and calculates values for several video signals with different formats. This table can be used as a handy quick reference to see the relationship between resolution and bandwidth.
Table: Performance Requirements for Various Video Standards













Horizontal Visual Resolution (TV Lines) 












Total Horizontal Active Pixels 












Total Vertical Active Lines 












Total Active Pixels per Frame (k) 












Aspect Ratio 












Ratio of Total to Active Horizontal Pixels 












Total Horizontal Pixels 












Ratio of Total to Active Vertical Lines 












Total Vertical Scan Lines 












Scan Method Interlaced (I)/ Progressive (P) 












Frame Rate (Hz) 












H Rate (kHz) 












Pixel Rate (Mp/s) 












Max Signal BW 












BW(3B) Nominal for 0.5dB flatness (Mhz) 












BW(3B) Nominal for 0.1dB flatness (Mhz) 












Slew Rate Nominal (V/µs) 












CircuitBandwidth and SlewRate Requirements
The circuits that process video signals need to have more bandwidth than the actual bandwidth of the processed signal to minimize the degradation of the signal and the resulting loss in picture quality. The amount the circuit bandwidth needs to exceed the highest frequency in the signal is a function of the quality desired. To calculate this, we assume a singlepole response and use the following equation:H(f)(dB) = 20log(1/(1+(BW_{S}/BW_{3dB})²)^{.5})Rearranging and solving for the 0.1dB and the 0.5dB attenuation points, we get the following:
BW_{3dB min} = BW_{S} (0.1db) × 6.55 EQ 3Where:
BW_{3dBmin} = BW_{S}(0.5db) × 2.86 EQ 4
BW_{3dB} = the minimum 3db bandwidth required for the circuit
From equations 3 and 4, if you want to keep the signal attenuation to less than 0.1dB, the circuit needs to have a minimum bandwidth that's about six and a half times' the highest frequency in the signal. If you can tolerate 0.5dB attenuation, it needs to be only about three times. To account for normal variations in the bandwidth of integrated circuits, it is recommended that the results from equations 3 and 4 be multiplied by a factor of 1.5. This will ensure that the attenuation performance is met over worstcase conditions. In equation mode, it is expressed as follows:
BW_{3dB nominal} = BW_{3dBmin} × 1.5 EQ 5In addition to bandwidth, the circuits must slew fast enough to faithfully reproduce the video signal. The equation for the minimum slew rate is as follows:
SR_{MIN} = 2 × pi × BW_{S }× VpeakSubstituting and simplifying,
SR_{MIN} = BW_{S} × 6.386 EQ 6For optimum performance, it is necessary to specify a slew rate larger than that given by equation 6. This is because some distortion can occur as the frequency of the signal approaches the slewrate limit. This can introduce frequency distortion, which will degrade the picture quality. Multiplying the equation 6 result by a factor of at least two or three will ensure that the distortion is minimized.
In equation form:
SR_{nominal} = SR_{MIN} × 2 EQ 7As an example, let's assume we have a standard NTSC video signal and the following requirements:
V_{LT} = 525
TVL = 346
AR = 1.3333
K_{H} = 1.17
F_{R} = 29.94
K_{V} = 1.09
Using equation 1, we calculate a maximum signal bandwidth (BW_{S}) of about 4.2MHz. This is the highest frequency in the signal. Now let's assume that we need less than 0.1dB attenuation. Using equation 3, we calculate the minimum signal bandwidth necessary to be 27.5MHz. Using equation 5, to account for variations, gives 41.3MHz. This is the circuit 3dB bandwidth required to achieve our desired resolution and maintain the signal quality.
The last calculation we need to make for our example is the minimum slewrate requirement. Using equations 6 and 7 and plugging in the 4.2MHz value for BW_{S}, we see that we will need at least a slew rate of 52V/µs and a more desirable value of 80V/µs.