## Do I Really Need a 4K (or 8K!) TV?

The short answer is, no and yes. Some analysts will have you believe that “8K TV blows 4K away,” and that might suggest that you at least want a 4K TV.  The reality, as it comes to electronics and perception, is more complicated.

One might assume that higher resolution always makes a picture better, because the pixels get smaller and smaller, to the point where you don’t see them anymore.  But the human visual system — your eyes — has a finite capacity, and once you exceed this, any other “improvement” is wasted, because it just won’t be seen.

Here’s why (warning, geometry involved):

The term “20/20 vision” is defined as the ability to just distinguish features that subtend one-arc-minute of angle (one-sixtieth of a degree). In other words, objects at a certain distance can only be resolved as separate objects if the objects are a certain distance apart.

Using trigonometry, this works out to be about 1/32″ as the smallest separation a person with 20/20 vision can see at a distance of ten feet. We can use the same math to show that the “optimum” distance from which to observe an HD (1080-line) display (i.e., where a 20/20 observer can just resolve the pixels) is about 3 times the picture height.

On a 1080-line monitor with a 15” diagonal, this works out to an optimum viewing distance of just under two feet; with a 42” display, it’s about five-and-a-half feet. Sitting closer than this means the pixels will become visible; sitting further means that the resolution is “wasted.”  Keep in mind, also, that most people sit about 9 feet away from the TV, what is sometimes called the “Lechner distance,” after a well-known TV systems researcher.

Of course, these numbers (and others produced by various respectable organizations) are based on subjective evaluation of the human visual system, and different observers will show different results, especially when the target applications vary.  Nonetheless, the “three picture heights” rule has survived critical scrutiny for several decades, and we haven’t seen a significant deviation in practice.

At 4K, the optimum distance becomes 1.6 picture-heights: at the same 1080-display viewing distance of 5.5 feet, one needs an 84”-diagonal display (7 feet), which is already available. For these reasons, some broadcasters believe that 4K is not a practical viewing format, since displaying 4K images would require viewing at 2.5 picture-heights to match normal human visual acuity.

At 8K, the numbers become absurd for the typical viewer: 0.7 picture heights, or a 195″ diagonal (16 feet) at a 5.5-foot distance.  With a smaller display, or at a larger distance, the increased resolution is completely invisible to the viewer: that means wasted pixels (and money).  Because such a display is very large (and thus very expensive), the 105-degree viewing angle it would subtend at the above viewing distance approaches a truly immersive and lifelike experience for a viewer — but how many people would put such a beast in their home?

From a production perspective, 4K does make some sense, because an environment that captures all content in 4K, and then processes this content in a 1080p workflow for eventual distribution, will produce archived material at a very high intrinsic quality.  Of course, there’s a cost associated with that, too.

But there are two other reasons why one might be persuaded to upgrade their HDTV:  HDR (High Dynamic Range) and HFR (High Frame Rate).  Briefly, HDR increases the dynamic range of video from about 6 stops (64:1) to more than 200,000:1 or 17.6 stops, making the detail and contrast appear closer to that of reality.  HFR increases the frame rate from the currently-typical 24, 30 or 60 fps to 120 fps.  And these other features make a much more recognizable improvement in pictures — at almost any level of eyesight.  But that’s another story.

agc

## There’s No Such Thing as RMS Power!

This is one of my engineering pet peeves — I keep running into students and (false) advertisements that describe a power output in “RMS watts.”  The fact is, such a construct, while mathematically possible, has no meaning or relevance in engineering.  Power is measured in watts, and while the concepts of average and peak watts is tenable, “RMS power” is a fallacy.  Here’s why.

The power dissipated by a resistive load is equal to the square of the voltage across the load, divided by the resistance of the load.  Mathematically, this is expressed as [Eq.1]:

$$\large P=\frac{V^{2}}{R}$$

where P is the power in watts, V is the voltage in volts, and R is the resistance in ohms.  When we have a DC signal, calculating the power in the load is straightforward.  The complication arises when we have a time-varying signal, such as an alternating current (AC), e.g, an audio signal or an RF signal.  In the case of power, the most elementary time-varying function involved is the sine function.

When measuring the power dissipated in a load carrying an AC signal, we have different ways of measuring that power.  One is the instantaneous or time-varying power, which is Equation 1 applied all along the sinusoid as a time-varying function.  (We will take r = 1 here, as a way of simplifying the discussion; in practice, we would use an appropriate value, e.g., 50Ω in the case of an RF load.)

In Figure 1, the dotted line (green) trace is our 1-volt (peak) sinusoid. (The horizontal axis is in degrees.) The square of this function (the power as a function of time) is the dark blue trace, which is essentially a “raised cosine” function.  Since the square is always a positive number, we see that the power as a function of time rises and falls as a sinusoid, at twice the frequency of the original voltage.  This function itself has relatively little use in most applications.

Another quantity is the peak power, which is simply Equation 1 above, where V is taken to be the peak value of the sinusoid, in this case, 1.  This is also known as peak instantaneous power (not to be confused with peak envelope power, or PEP).  The peak instantaneous power is useful to understand certain limitations of electronic devices, and is expressed as follows:

$$\large P_{pk}=\frac{V^{2}_{pk}}{R}$$

A more useful quantity is the average power, which will provide the equivalent heating factor in a resistive device.  This is calculated by taking the mean of the square of the voltage signal, divided by the resistance. Since the sinusoidal power function is symmetric about its vertical midpoint, simple inspection (see Figure 1 again) tells us that the mean value is equal to one-half of the peak power [Eq.2]:

$$\large P_{avg}=\frac{P_{pk}}{2}=\frac{V^{2}_{pk}/R}{2}$$

which in this case is equal to 0.5.  We can see this in Figure 1, where the average of the blue trace is the dashed red trace.  Thus, our example of a one-volt-peak sinusoid across a one-ohm resistor will result in an average power of 0.5 watts.

Now the concept of “RMS” comes in, which stands for “root-mean-square,” i.e., the square-root of the mean of the square of a function.  (The “mean” is simply the average.) The purpose of RMS is to present a particular statistical property of that function.  In our case, we want to associate a “constant” value with a time-varying function, one that provides a way of describing the “DC-equivalent heating factor” of a sinusoidal signal.

Taking the square-root of  V2pk/2 therefore provides us with the root-mean-square voltage (not power) across the resistor; in this example, that means that the 1-volt (peak) sinusoid has an RMS voltage of

$$\large V_{rms}=\sqrt{\frac{V^{2}_{pk}}{2}}=\frac{V_{pk}}{\sqrt{2}}\approx 0.7071$$

Thus, if we applied a DC voltage of 0.7071 volts across a 1Ω resistor, it would consume the same power (i.e., dissipate the same heat) as an AC voltage of 1 volt peak.  (Note that the RMS voltage does not depend on the value of the resistance, it is simply related to the peak voltage of the sinusoidal signal.) Plugging this back into Eq. 2 then gives us:

$$\large P_{avg}=\frac{V^{2}_{rms}}{R}$$

Note the RMS voltage is used to calculate the average power. As a rule, then, we can calculate the RMS voltage of a sinusoid this way:

$$\large V_{rms} \approx 0.7071 \cdot V_{pk}$$

Graphically, we can see this in Figure 2:

The astute observer will note that 0.7071 is the value of sin(45°) to four places. This is not a coincidence, but we leave it to the reader to figure out why.  Note that for more complex signals, the 0.7071 factor no longer holds.  A triangle wave, for example, yields Vrms ≈ 0.5774 · Vpk , where 0.5774 is the value of tan(30°) to four places.

For those familiar with calculus, the root-mean-square of an arbitrary function is defined as:

$$\large F_{rms} = \sqrt{\frac{1}{T_{2}-T_{1}}\int_{T_{1}}^{T_{2}}[f(t)]^{2}\, dt}$$

Replacing f(t) with sin(t) (or an appropriate function for a triangle wave) will produce the numerical results we derived above.